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Organic Structures from Spectra Fifth Edition
i
ii
Organic Structures from Spectra Fifth Edition L. D. Field University o f New South Wales, Australia
S. Sternhell University o f Sydney, Australia
J. R. Kalman University o f Technology Sydney, Australia
iii
This edition first published 2013 © 2013 John Wiley & Sons, Ltd. Registered Office John Wiley & Sons Ltd, The Atrium, Southern Gate, Chichester, West Sussex, PO19 8SQ, United Kingdom For details of our global editorial offices, for customer services and for information about how to apply for permission to reuse the copyright material in this book please see our website at www.wiley.com. The right of the author to be identified as the author of this work has been asserted in accordance with the Copyright, Designs and Patents Act 1988. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, except as permitted by the UK Copyright, Designs and Patents Act 1988, without the prior permission of the publisher. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic books. Designations used by companies to distinguish their products are often claimed as trademarks. All brand names and product names used in this book are trade names, service marks, trademarks or registered trademarks of their respective owners. The publisher is not associated with any product or vendor mentioned in this book. This publication is designed to provide accurate and authoritative information in regard to the subject matter covered. It is sold on the understanding that the publisher is not engaged in rendering professional services. If professional advice or other expert assistance is required, the services of a competent professional should be sought. The publisher and the author make no representations or warranties with respect to the accuracy or completeness of the contents of this work and specifically disclaim all warranties, including without limitation any implied warranties of fitness for a particular purpose. This work is sold with the understanding that the publisher is not engaged in rendering professional services. The advice and strategies contained herein may not be suitable for every situation. In view of ongoing research, equipment modifications, changes in governmental regulations, and the constant flow of information relating to the use of experimental reagents, equipment, and devices, the reader is urged to review and evaluate the information provided in the package insert or instructions for each chemical, piece of equipment, reagent, or device for, among other things, any changes in the instructions or indication of usage and for added warnings and precautions. The fact that an organization or Website is referred to in this work as a citation and/or a potential source of further information does not mean that the author or the publisher endorses the information the organization or Website may provide or recommendations it may make. Further, readers should be aware that Internet Websites listed in this work may have changed or disappeared between when this work was written and when it is read. No warranty may be created or extended by any promotional statements for this work. Neither the publisher nor the author shall be liable for any damages arising herefrom. Library of Congress Cataloging-in-Publication Data applied for HB ISBN: 9781118325452 PB ISBN: 9781118325490
A catalogue record for this book is available from the British Library. Typeset in 12/18pt Times New Roman by Aptara Inc., New Delhi, India Printed in Singapore by Markono Print Media Pte Ltd
iv
CONTENTS PREFACE LIST OF TABLES LIST OF FIGURES 1 INTRODUCTION 1.1 1.2 1.3 1.4 1.5 1.6
2
ULTRAVIOLET (UV) SPECTROSCOPY 2.1 2.2 2.3 2.4 2.5 2.6 2.7
3
ABSORPTION RANGE AND THE NATURE OF IR ABSORPTION EXPERIMENTAL ASPECTS OF INFRARED SPECTROSCOPY GENERAL FEATURES OF INFRARED SPECTRA IMPORTANT IR CHROMOPHORES
MASS SPECTROMETRY 4.1 4.2 4.3 4.4 4.5 4.6
5
BASIC INSTRUMENTATION THE NATURE OF ULTRAVIOLET SPECTROSCOPY QUANTITATIVE ASPECTS OF ULTRAVIOLET SPECTROSCOPY CLASSIFICATION OF UV ABSORPTION BANDS SPECIAL TERMS IN ULTRAVIOLET SPECTROSCOPY IMPORTANT UV CHROMOPHORES THE EFFECT OF SOLVENTS
INFRARED (IR) SPECTROSCOPY 3.1 3.2 3.3 3.4
4
GENERAL PRINCIPLES OF ABSORPTION SPECTROSCOPY CHROMOPHORES DEGREE OF UNSATURATION CONNECTIVITY SENSITIVITY PRACTICAL CONSIDERATIONS
IONISATION PROCESSES INSTRUMENTATION MASS SPECTRAL DATA REPRESENTATION OF FRAGMENTATION PROCESSES FACTORS GOVERNING FRAGMENTATION PROCESSES EXAMPLES OF COMMON TYPES OF FRAGMENTATION
NUCLEAR MAGNETIC RESONANCE(NMR) SPECTROSCOPY 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10
THE PHYSICS OF NUCLEAR SPINS AND NMR INSTRUMENTS CONTINUOUS WAVE (CW) NMR SPECTROSCOPY FOURIER-TRANSFORM (FT) NMR SPECTROSCOPY THE NUCLEAR OVERHAUSER EFFECT (NOE) CHEMICAL SHIFT IN 1H NMR SPECTROSCOPY SPIN-SPIN COUPLING IN 1H NMR SPECTROSCOPY ANALYSIS OF 1H NMR SPECTRA CHANGING THE MAGNETIC FIELD IN NMR SPECTROSCPY RULES FOR SPECTRAL ANALYSIS OF FIRST ORDER SPECTRA CORRELATION OF 1H - 1H COUPLING CONSTANTS WITH STRUCTURE
vii xi xiii 1 1 3 3 4 5 5 7 7 8 8 9 10 10 14 15 15 16 16 17
21 21 23 24 29 30 30 34 34 39 40 41 42 53 56 58 59 66
V
Contents
6 13C NMR SPECTROSCOPY 6.1 6.2 6.3
COUPLING AND DECOUPLING IN 13C NMR SPECTRA DETERMINING 13C SIGNAL MULTIPLICITY USING DEPT SHIELDING AND CHARACTERISTIC CHEMICAL SHIFTS IN 13C NMR SPECTRA
7 2-DIMENSIONAL NMR SPECTROSCOPY 7.1 7.2
7.3 7.4 7.5
COSY (CORRELATION SPECTROSCOPY) THE HSQC (HETERONUCLEAR SINGLE QUANTUM CORRELATION) OR HSC (HETERONUCLEAR SHIFT CORRELATION) SPECTRUM HMBC (HETERONUCLEAR MULTIPLE BOND CORRELATION) NOESY (NUCLEAR OVERHAUSER EFFECT SPECTROSCOPY) TOCSY (TOTAL CORRELATION SPECTROSCOPY)
8 MISCELLANEOUS TOPICS 8.1 8.2 8.3 8.4 8.5
SOLVENTS FOR NMR SPECTROSCOPY SOLVENT INDUCED SHIFTS DYNAMIC PROCESSES IN NMR ■THE NMR TIME-SCALE THE EFFECT OF CHIRALITY THE NMR SPECTRA OF "OTHER NUCLEI"
9 DETERMINING THE STRUCTURE OF ORGANIC COMPOUNDS FROM SPECTRA 9.1 9.2
71 72 75
81 83 84
86 91 92
94 94 95 96 98 99
100 102 103
10 PROBLEMS
111
10.1 10.2 10.3 10.4
111 397 407 467
INDEX
VI
SOLVING PROBLEMS WORKED EXAMPLES
71
SPECTROSCOPIC IDENTIFICATION OF ORGANIC COMPOUNDS THE ANALYSIS OF MIXTURES PROBLEMS IN 2-DIMENSIONAL NMR NMR SPECTRAL ANALYSIS
493
PREFACE
The derivation of structural information from spectroscopic data is an integral part of Organic Chemistry courses at all Universities. At the undergraduate level, the principal aim of courses in organic spectroscopy is to teach students to solve simple structural problems efficiently by using combinations of the major techniques (UV, IR, NMR and MS). Over a period more than 30 years, we have evolved courses at the University of Sydney and at the University of New South Wales, which achieve this aim quickly and painlessly. The text is tailored specifically to the needs and philosophy of these courses. As we believe our approach to be successful, we hope that it may be of use in other institutions. The courses has been taught at the beginning of the third year, at which stage students have completed an elementary course of Organic Chemistry in first year and a mechanistically-oriented intermediate course in second year. Students have also been exposed, in their Physical Chemistry courses, to elementary spectroscopic theory, but are, in general, unable to relate the theory to actually solving spectroscopic problems. We have delivered courses of about 9 lectures outlining the basic theory, instrumentation and the structure-spectra correlations of the major spectroscopic techniques. The text of this book broadly corresponds to the material presented in the 9 lectures. The treatment is both elementary and condensed and, not surprisingly, the students have great difficulties in solving even the simplest problems at this stage. The lectures are followed by a series of 2-hour problem solving seminars with 5 to 6 problems being presented per seminar. At the conclusion of the course, the great majority of the class is quite proficient and has achieved a satisfactory level of understanding of all methods used. Clearly, the real teaching is done during the hands-on problem seminars, which are organised in a manner modelled on that which we first encountered at the E.T.H. Zurich. The class (typically 60 - 100 students, attendance is compulsory) is seated in a large lecture theatre in alternate rows and the problems for the day are identified. The students are permitted to work either individually or in groups and may use any written or printed aids they desire. Students solve the problems on their individual copies of this book thereby transforming it into a set of worked examples and most students voluntarily complete many more problems than are set. Staff (generally 4 or 5) wander around giving help and tuition as needed - the empty alternate rows of
vii
Preface
seats make it possible to speak to each student individually. When an important general point needs to be made, the staff member in charge gives a very brief exposition at the board. There is a 1½ hour examination consisting essentially of 4 problems from the book and the results are in general very satisfactory. Moreover, the students themselves find this a rewarding course since the practical skills acquired are obvious to them. Solving these real puzzles is also addictive - there is a real sense of achievement, understanding and satisfaction, since the challenge in solving the graded problems builds confidence even though the more difficult examples are quite demanding. Our philosophy can be summarised as follows: (a)
Theoretical exposition must be kept to a minimum, consistent with gaining of an understanding of the parts of the technique actually used in solving the problems. Our experience indicates that both mathematical detail and description of advanced techniques merely confuse the average student.
(b)
The learning of data must be kept to a minimum. We believe that it is more important to learn to use a restricted range of data well rather than to achieve a nodding acquaintance with more extensive sets of data.
(c)
Emphasis is placed on the concept of identifying "structural elements" and the logic needed to produce a structure out of the structural elements.
We have concluded that the best way to learn how to obtain "structures from spectra" is to practise on simple problems. This book was produced principally to assemble a suitable collection of problems for that purpose. Problems 1-282 are of the standard “structures from spectra” type and are arranged roughly in order of increasing difficulty. A number of problems deal with related compounds (sets of isomers) which differ mainly in symmetry or the connectivity of the structural elements and are ideally set together. The sets of related examples include: problems 3 and 4; 19 and 20; 31 and 32; 42 and 43; 44, 45 and 46; 47, 48 and 49; 50 and 51; 61, 62 and 63; 64, 65 and 6 6 ; 81 and 82; 84 and 85; 99, 100, 101 and 102; 107 and 108; 1 10, 111, 112 and 113; 114 and 115; 118, 119 and 120; 122 and 123; 127 and 128; 139, 140, 141, 142 and 143; 155, 156, 157, 158, 159 and 160; 179 and 180; 181 and 182; 185 and 186; 215 and 216; 226 and 227; 235, 236 and 237; 276 and 277. A further group of problems offer practice in the analysis of proton NMR spectra: 19, 20, 29, 37, 58, 75, 79, 90, 92, 93, 94, 99, 101, 123, 137, 146, 159, 163, 164, 183, 187, 192, 195, 205, 208, 236, 237, 238, 239, 248, 250, 251, 252 and 260.
viii
Preface
A number of problems (195, 196, 197, 198, 230, 231, 260, 264, 265, 268, 271, 274 and 275) exemplify complexities arising from the presence of chiral centres, or from restricted rotation about peptide bonds (128, 162 and 262), while some problems deal with structures of compounds of biological, environmental, or industrial significance (22, 23, 36, 8 6 , 95, 127, 131, 132, 144, 153, 162, 164, 197, 204, 220, 259, 260, 261, 263, 264, 265, 267, 272, 273, 274 and 275). Problems 283-288 are again structures from spectra, but with the data presented in a textual form such as might be encountered when reading the experimental section of a paper or report. Problems 289-296 deal with the use of NMR spectroscopy for quantitative analysis and for the analysis of mixtures of compounds. Problems 297-323 represent a considerably expanded set of problems dealing with the interpretation of two-dimensional NMR spectra and are a series of graded exercises utilising COSY, NOESY, C-H Correlation, HMBC and TOCSY spectroscopy as aids to spectral analysis and as tools for identifying organic structures from spectra. Problems 324-346 deal specifically with more detailed analysis of NMR spectra, which tends to be a stumbling block for many students. In Chapter 9, there are also two worked solutions (to problems 96 and 127) as an illustration of a logical approach to solving problems. However, with the exception that we insist that students perform all routine measurements first, we do not recommend a mechanical attitude to problem solving - intuition has an important place in solving structures from spectra as it has elsewhere in chemistry.
Bona fide instructors may obtain a list of solutions (at no charge) by writing to the authors or EMAIL: [email protected] or FAX: (61-2)-9385-8008
We wish to thank Dr Alison Magill, and Dr Hsiu Lin Li in the School of Chemistry at the University of New South Wales and Dr Ian Luck at the University of Sydney who helped to assemble the many additional samples and spectra in the 4th and 5th editions of this book. Thanks are also due to the many graduate students and research associates who, over the years, have supplied us with many of the compounds used in the problems. L. D. Field S. Sternhell J. R. Kalman September 2012
ix
X
LIST OF TABLES Table 2.1
The Effect of Extended Conjugation on UV Absorption
11
Table 2.2
UV Absorption Bands in Common Carbonyl Compounds
12
Table 2.3
UV Absorption Bands in Common Benzene Derivatives
13
Table 3.1
Carbonyl (C =0) IR Absorption Frequencies in Common Functional Groups
18
Table 3.2
Characteristic IR Absorption Frequencies for Functional Groups
19
Table 3.3
Common IR Absorption Frequencies in the Region 1900 - 2600 cm '1
20
Table 4.1
Accurate Masses of Selected Isotopes
25
Table 4.2
Common Fragments and their Masses
27
Table 5.1
Resonance Frequencies of ]H and 13C Nuclei in Magnetic Fields of Different Strengths
36
Table 5.2
Typical
45
Table 5.3
Typical lR Chemical Shift Ranges in Organic Compounds
Table 5.4
Chemical Shift Values in Selected Organic Compounds
Chemical Shifts (8) for Protons in Common Alkyl Derivatives
45 46
Table 5.5
Approximate lH Chemical Shifts (8) for Olefinic Protons C=C-H in ppm
48
Table 5.6
Approximate lH Chemical Shifts (8) for Aromatic Protons in Benzene Derivatives Ph-X in ppm Relative to Benzene at 8 7.26 ppm
49
Table 5.7
]H Chemical Shifts (8) in some Polynuclear Aromatic Compounds and Heteroaromatic Compounds
49
Table 5.8
Typical lR -
54
Table 5.9
Relative Line Intensities for Simple Multiplets
54
Table 6.1
The Number of Aromatic 13C Resonances in Benzenes with Different Substitution Patterns
74
Table 6.2
Typical 13C Chemical Shift Values in Selected Organic Compounds
75
Table 6.3
Typical 13C Chemical Shift Ranges in Organic Compounds
76
Table 6.4
13C Chemical Shifts (8) for sp3 Carbons in Alkyl Derivatives
78
Table 6.5
13C Chemical Shifts (8) for sp2 Carbons in Vinyl Derivatives CH2=CH-X
78
Table 6.6
13C Chemical Shifts (8) for sp Carbons in Alkynes: X-C=C-Y
79
Table 6.7
Approximate 13C Chemical Shifts (8) for Aromatic Carbons in Benzene Derivatives Ph-X in ppm relative to Benzene at 8 128.5 ppm
80
Table 6.8
Characteristic 13C Chemical Shifts (8) in some Polynuclear Aromatic Compounds and Heteroaromatic Compounds
80
Table 8.1
]H and 13C Chemical Shifts for Common NM R Solvents
95
Coupling Constants
xi
xii
LIST OF FIGURES Figure 1.1
Schematic Absorption Spectrum
1
Figure 1.2
Definition o f a Spectroscopic Transition
2
Figure 2.1
Schematic Representation o f an IR or UV Spectrometer
7
Figure 2.2
Definition of Absorbance (A)
9
Figure 4.1
Schematic Diagram of an Electron-Impact Mass Spectrometer
23
Figure 4.2
Relative Intensities of the Cluster of Molecular Ions for Molecules Containing Combinations of Bromine and Chlorine Atoms
29
Figure 5.1
A Spinning Charge Generates a Magnetic Field and Behaves Like a Small Magnet
34
Figure 5.2
Schematic Representation of a CW NM R Spectrometer
39
Figure 5.3
Time Domain and Frequency Domain NM R Spectra
40
Figure 5.4
Approximate ]H Chemical Shift Ranges for Protons in Organic Compounds
47
Figure 5.5
Shielding/deshielding Zones for Common Non-aromatic Functional Groups
51
Figure 5.6
D20 Exchange in the *H Spectrum of 1-Propanol
52
Figure 5.7
Characteristic Multiplet Patterns for Common Organic Fragments
55
Figure 5.8
A Portion o f the solution in CC14)
61
Figure 5.9
The 60 MHz *H NMR Spectrum of a 4-Spin AMX2 Spin System
NMR Spectrum o f Styrene Epoxide (100 MHz as a 5%
62
Figure 5.10 Simulated lH NMR Spectra of a 2-Spin System as the Ratio Av/J, is Varied from 10.0 to 0.0
63
Figure 5.11
Selective Decoupling in a Simple 4-Spin System
65
Figure 5.12
Characteristic Aromatic Splitting Patterns in the !H NM R spectra for some Diand Tri-substituted Benzenes
69
Figure 5.13
jH NM R Spectrum ofj^-Nitrophenylacetylene (200 MHz as a 10% solution in CDC13)
70
Figure 6.1
13C NM R Spectra of Methyl Cyclopropyl Ketone (CDC13 Solvent, 100 MHz). (a) with Broad Band Decoupling of *H; (b) DEPT Spectrum (c) with no Decoupling of *H
73
Figure 6.2
Approximate 13C Chemical Shift Ranges for Carbon Atoms in Organic Compounds
77
Figure 7.1
]H COSY Spectrum of 1-Iodobutane
83
Figure 7.2
]H - 13C HSQC Spectrum of 1-Iodobutane
85
Figure 7.3
]H - 13C HMBC Spectrum of 1-Iodobutane
87
Figure 7.4
*H - 13C HMBC Spectrum of 2-Bromophenol
89
Figure 7.5
]H NOESY Spectrum of (3-Butyrolactone
91
Figure 7.6
]H TOCSY Spectrum of Butyl Ethyl Ether
93
Figure 8.1
Schematic NMR Spectra of Two Exchanging Nuclei
96
Figure 8.2
]H NM R Spectrum of the Aliphatic Region o f Cysteine
99
xiv
1 INTRODUCTION
1.1
GENERAL PRINCIPLES OF ABSORPTION SPECTROSCOPY
The basic principles of absorption spectroscopy are summarised below. These are most obviously applicable to UV and IR spectroscopy and are simply extended to cover NMR spectroscopy. Mass Spectrometry is somewhat different and is not a type of absorption spectroscopy. Spectroscopy is the study of the quantised interaction of energy (typically electromagnetic energy) with matter. In Organic Chemistry, we typically deal with molecular spectroscopy i.e. the spectroscopy of atoms that are bound together in molecules. A schematic absorption spectrum is given in Figure 1.1. The absorption spectrum is a plot of absorption of energy (radiation) against its wavelength (X) or frequency (v).
AE
Figure 1.1
Schematic Absorption Spectrum
Organic Structures from Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
Chapter 1 Introduction
An absorption band can be characterised primarily by two parameters: (a)
the wavelength at which maximum absorption occurs
(b)
the intensity of absorption at this wavelength compared to base-line (or background) absorption
A spectroscopic transition takes a molecule from one state to a state of a higher energy. For any spectroscopic transition between energy states (e.g. E, and E2 in Figure 1.2), the change in energy (AE) is given by: AE = hv where h is the Planck's constant and v is the frequency of the electromagnetic energy absorbed. Therefore v oc AE.
A
----------------
I
t AE = E2 - Ei
Energy
-------Figure 1.2
e2
Ei
Definition of a Spectroscopic Transition
It follows that the x-axis in Figure 1.1 is an energy scale, since the frequency, wavelength and energy of electromagnetic radiation are interrelated: vX = c (speed of light) X=±-
v
iA 1 , oc----
AE A spectrum consists of distinct bands or transitions because the absorption (or emission) of energy is quantised. The energy gap of a transition is a molecular property and is characteristic o f molecular structure. The y-axis in Figure 1.1 measures the intensity of the absorption band and this depends on the number of molecules observed (the Beer-Lambert Law) and the probability of the transition between the energy levels. The absorption intensity is also a molecular property and both the frequency and the intensity of a transition can provide structural information.
2
Chapter 1 Introduction
1.2
CHROMOPHORES
In general, any spectral feature, i.e. a band or group of bands, is due not to the whole molecule, but to an identifiable part of the molecule, which we loosely call a chromophore. A chromophore may correspond to a functional group (e.g. a hydroxyl group or the double bond in a carbonyl group). However, it may equally well correspond to a single atom within a molecule or to a group of atoms (e.g. a methyl group) which is not normally associated with chemical functionality. The detection of a chromophore permits us to deduce the presence of a structural fragment or a structural element in the molecule. The fact that it is the chromophores and not the molecules as a whole that give rise to spectral features is fortunate, otherwise spectroscopy would only permit us to identify known compounds by direct comparison of their spectra with authentic samples. This "fingerprint" technique is often useful for establishing the identity of known compounds, but the direct determination of molecular structure building up from the molecular fragments is far more powerful.
1.3
DEGREE OF UNSATURATION
Traditionally, the molecular formula of a compound was derived from elemental analysis and its molecular weight which was determined independently. The concept of the degree of unsaturation of an organic compound derives simply from the tetravalency of carbon. For a non-cyclic hydrocarbon (i.e. an alkane) the number of hydrogen atoms must be twice the number of carbon atoms plus two, any “deficiency” in the number of hydrogens must be due to the presence of unsaturation, i.e. double bonds, triple bonds or rings in the structure. The degree of unsaturation can be calculated from the molecular formula for all compounds containing C, H, N, O, S or the halogens. There are 3 basic steps in calculating the degree of unsaturation: Step 1 - take the molecular formula and replace all halogens by hydrogens Step 2 - omit all of the sulfur or oxygen atoms Step 3 - for each nitrogen, omit the nitrogen and omit one hydrogen
3
Chapter 1 Introduction
After these 3 steps, the molecular formula is reduced to CnHmand the degree of unsaturation is given by: Degree of Unsaturation = n - ~
+1
The degree of unsaturation indicates the number of n bonds or rings that the compound contains. For example, a compound whose molecular formula is C4H9NO 2 is reduced to C4H8 which gives a degree of unsaturation of 1 and this indicates that the molecule must have one n bond or one ring. Note that any compound that contains an aromatic ring always has a degree of unsaturation greater than or equal to 4 , since the aromatic ring contains a ring plus three n bonds. Conversely, if a compound has a degree of unsaturation greater than 4, one should suspect the possibility that the structure contains an aromatic ring.
1.4
CONNECTIVITY
Even if it were possible to identify sufficient structural elements in a molecule to account for the molecular formula, it may not be possible to deduce the structural formula from a knowledge of the structural elements alone. For example, it could be demonstrated that a substance of molecular formula C3H 50C1 contains the structural elements: -
ch 3 — Cl
\
/
c=o
— ch 2and this leaves two possible structures:
C H o-C -C H 2-C I d II
CHo-CH2- C - C I and
J
ii
O
0
1
2
Not only the presence of various structural elements, but also their juxtaposition, must be determined to establish the structure of a molecule. Fortunately, spectroscopy often gives valuable information concerning the connectivity of structural elements and in the above example it would be very easy to determine whether there is a
4
Chapter 1 Introduction
ketonic carbonyl group (as in 1) or an acid chloride (as in 2). In addition, it is possible to determine independently whether the methyl (-CH3) and methylene (-CH2-) groups are separated (as in 1) or adjacent (as in 2).
1.5
SENSITIVITY
Sensitivity is generally taken to signify the limits of detectability of a chromophore. Some methods (e.g. [H NMR) detect all chromophores accessible to them with equal sensitivity while in other techniques (e.g. UV) the range of sensitivity towards different chromophores spans many orders of magnitude. In terms of overall sensitivity, i.e. the amount of sample required, it is generally observed that: MS > UV > IR > >H NMR > >3C NMR but considerations of relative sensitivity toward different chromophores may be more important.
1.6
PRACTICAL CONSIDERATIONS
The 5 major spectroscopic methods (MS, UV, IR, ]H NMR and ,3C NMR) have become established as the principal tools for the determination of the structures of organic compounds, because between them they detect a wide variety of structural elements. The instrumentation and skills involved in the use of all five major spectroscopic methods are now widely spread, but the ease of obtaining and interpreting the data from each method under real laboratory conditions varies. In very general terms: (a)
While the cost of each type of instrumentation differs greatly (NMR instruments cost between $50,000 and several million dollars), as an overall guide, MS and NMR instruments are much more costly than UV and IR spectrometers. With increasing cost goes increasing difficulty in maintenance and the required operator expertise, thus compounding the total outlay.
(b)
In terms of ease o f usage for routine operation, most UV and IR instruments are comparatively straightforward. NMR Spectrometers are also common as “hands-on” instruments in most chemistry laboratories and the users require routine training and a degree of basic computer literacy. Similarly some Mass Spectrometers are now designed to be used by researchers as “hands-on” routine instruments. However, the more advanced NMR Spectrometers and most Mass
5
Chapter 1 Introduction
Spectrometers are sophisticated instruments that are usually operated and maintained by specialists. (c)
The scope of each spectroscopic method can be defined as the amount of useful information it provides. This is a function of the total amount of information obtainable and also how difficult the data are to interpret. The scope of each method varies from problem to problem, and each method has its aficionados and specialists, but the overall utility undoubtedly decreases in the order: NMR > MS > IR > UY with the combination of ’H and l3C NMR providing the most useful information.
(d)
The theoretical background needed for each method varies with the nature of the experiment, but the minimum overall amount of theory needed decreases in the order: NMR » MS > UV « IR
6
2 ULTRAVIOLET (UV) SPECTROSCOPY
2.1
BASIC INSTRUMENTATION
Basic instrumentation for both UV and IR spectroscopy consists of an energy source, a sample cell, a dispersing device (prism or grating) and a detector, arranged as schematically shown in Figure 2.1.
dispersing device or prism
radiation source
Figure 2.1
detector
sample
Schematic Representation of an IR or UV Spectrometer
The drive of the dispersing device is synchronised with the x-axis of the recorder or fed directly to a computer, so that this indicates the wavelength of radiation reaching the detector. The signal from the detector is transmitted to the y-axis of the recorder or to a computer and this indicates how much radiation is absorbed by the sample at any particular wavelength. Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
7
Chapter 2 Ultraviolet Spectroscopy
In practice, double-beam instruments are used where the absorption of a reference cell, containing only solvent, is subtracted from the absorption of the sample cell. Double beam instruments also cancel out absorption due to the atmosphere in the optical path as well as the solvent. The energy source must be appropriate for the wavelengths of radiation being scanned. The materials from which the dispersing device and the detector are constructed must be as transparent as possible to wavelengths being scanned. For UV measurements, the cells and optical components are typically made of quartz and ethanol, hexane, water or dioxane are usually chosen as solvents.
2.2
THE NATURE OF ULTRAVIOLET SPECTROSCOPY
The term "UV spectroscopy" generally refers to electronic transitions occurring in the region of the electromagnetic spectrum (X in the range 200-380 nm) accessible to standard UV spectrometers. Electronic transitions are also responsible for absorption in the visible region (approximately 380-800 nm) which is easily accessible instrumentally but of less importance in the solution of structural problems, because most organic compounds are colourless. An extensive region at wavelengths shorter than ~ 200 nm ("vacuum ultraviolet") also corresponds to electronic transitions, but this region is not readily accessible with standard instruments. UV spectra used for determination of structures are invariably obtained in solution.
2.3
QUANTITATIVE ASPECTS OF ULTRAVIOLET SPECTROSCOPY
The y-axis of a UV spectrum may be calibrated in terms of the intensity of transmitted light (i.e. percentage of transmission or absorption), as is shown in Figure 2.2, or it may be calibrated on a logarithmic scale i.e. in terms of absorbance (A) defined in Figure 2.2. Absorbance is proportional to concentration and path length (the Beer-Lambert Law). The intensity of absorption is usually expressed in terms of molar absorbance or the molar extinction coefficient (s) given by:
where M is the molecular weight, C the concentration (in grams per litre) and / is the path length through the sample in centimetres.
Chapter 2 Ultraviolet Spectroscopy
Figure 2.2
Definition of Absorbance (A)
UV absorption bands (Figure 2.2) are characterised by the wavelength of the absorption maximum (/-max) and s. The values of s associated with commonly encountered chromophores vary between 10 and 105. For convenience, extinction coefficients are usually tabulated as logio(s) as this gives numerical values which are easier to manage. The presence of small amounts of strongly absorbing impurities may lead to errors in the interpretation of UV data.
2.4
CLASSIFICATION OF UV ABSORPTION BANDS
UV absorption bands have fine structure due to the presence of vibrational sub-levels, but this is rarely observed in solution due to collisional broadening. As the transitions are associated with changes of electron orbitals, they are often described in terms of the orbitals involved, e.g. a -> a* 71 —> K* n —» 7i* n
a*
where n denotes a non-bonding orbital, the asterisk denotes an antibonding orbital and ct and n have the usual significance. Another method of classification uses the symbols: B
(for benzenoid)
E
(for ethylenic)
R
(for radical-like)
K
(for conjugated - from the German "konjugierte")
9
Chapter 2 Ultraviolet Spectroscopy
A molecule may give rise to more than one band in its UV spectrum, either because it contains more than one chromophore or because more than one transition of a single chromophore is observed. However, UV spectra typically contain far fewer features (bands) than IR, MS or NMR spectra and therefore have a lower information content. The ultraviolet spectrum of acetophenone in ethanol contains 3 easily observed bands:
logio(s) O C.
(nm) CH,
acetophenone
2.5
Assignment
244
12,600
4.1
7t -> 71*
K
280
1,600
3.2
7T
71*
B
317
60
1.8
n
71*
R
SPECIAL TERMS IN UV SPECTROSCOPY
Auxochromes (auxiliary chromophores) are groups which have little UV absorption by themselves, but which often have significant effects on the absorption (both Xmax and s) of a chromophore to which they are attached. Generally, auxochromes are atoms with one or more lone pairs e.g. -OH, -OR, -NR2, -halogen. If a structural change, such as the attachment of an auxochrome, leads to the absorption maximum being shifted to a longer wavelength, the phenomenon is termed a bathochromic shift. A shift towards shorter wavelength is called a hypsochromic shift.
2.6
IMPORTANT UV CHROMOPHORES
Most of the reliable and useful data is due to relatively strongly absorbing chromophores (s > 2 00 ) which are mainly indicative of conjugated or aromatic systems. Examples listed below encompass most of the commonly encountered effects.
Chapter 2 Ultraviolet Spectroscopy
(1)
Dienes and Polyenes
Extension of conjugation in a carbon chain is always associated with a pronounced shift towards longer wavelength, and usually towards greater intensity (Table 2.1).
Table 2.1
The Effect of Extended Conjugation on UV Absorption
Alkene
165
10,000
4.0
CH3-CH2-CH=CH-CH2-CH3 (trans)
184
10,000
4.0
c h 2=ch-ch=ch 2
217
20,000
4.3
CH3-CH=CH-CH=CH2 (trans)
224
23,000
4.4
CH2=CH-CH=CH-CH=CH2 (trans)
263
53,000
4.7
CH3-(CH=CH)5-CH3 (trans)
341
126,000
5.1
X
II CM X
CM
logio(e)
o
e
o
^max(nm)
When there are more than 8 conjugated double bonds, the absorption maximum of polyenes is such that they absorb light strongly in the visible region of the spectrum. Empirical rules (Woodward's Rules) of good predictive value are available to estimate the positions of the absorption maxima in conjugated alkenes and conjugated carbonyl compounds. The stereochemistry and the presence of substituents also influence UV absorption by the diene chromophore. For example:
V x = 214nm
^max = 2 5 3 n m
s = 16,000
s = 8,000
logio(s) = 4.2
logio(s) = 3.9
11
Chapter 2 Ultraviolet Spectroscopy
(2)
Carbonyl compounds
All carbonyl derivatives exhibit weak (s < 100) absorption between 250 and 350 nm, and this is only of marginal use in determining structure. However, conjugated carbonyl derivatives always exhibit strong absorption (Table 2.2).
Table 2.2
UV Absorption Bands in Common Carbonyl Compounds
Compound
o
/
w
CO
X
o -x
Acetone
o
Acetaldehyde
Structure
C H 3 x
^0 c 1 ch3
c
c
1
1
CH3
4-Methylpent-3-en-2-one ch3
Cyclohex-2-en-1 -one
Benzoquinone
12
1
ch3
o=° 0 = = o
Sulfones
i o=s=o I
Sulfonamides and Sulfonate esters
— SOo-N^
Alcohols
— C-OH
^
\
— s o 2- o —
/
Ethers Alkyl fluorides
\
— C-OR /
\ — C -F
|
1
/
Alkyl chlorides Alkyl bromides Alkyl iodides
\
— C -C I / \
19
Chapter 3 Infrared Spectroscopy
Carbon-carbon double bonds in unconjugated alkenes usually exhibit weak to moderate absorptions due to C=C stretching in the range 1660-1640 cm'1. Disubstituted, trisubstituted and tetrasubstituted alkenes usually absorb near 1670 cm'1. The more polar carbon-carbon double bonds in enol ethers and enones usually absorb strongly between 1600 and 1700 cm-1. Alkenes conjugated with an aromatic ring absorb strongly near 1625 cm-1. (4)
Chromophores absorbing in the region between 1900 and 2600 cm-1- The
absorptions listed in Table 3.3 often yield useful information because, even though some are of only weak or medium intensity, they occur in regions largely devoid of absorption by other commonly occurring chromophores.
20
Functional group
Structure
v (cm '1)
Intensity
alkyne
— c= c—
2100-2300
weak to medium
nitrile
---- C = N
2215 -2280
medium
cyanate
---- O— C = N
2130-2270
strong
thiocyanate
---- S— C = N
2130-2175
medium
isocyanate
---- N = C — O
2200 - 2300
strong broad
isothiocyanate
---- N = C = S
2000 - 2200
strong
allene
II o
Table 3.3 Common IR Absorption Frequencies in the Region 1900 - 2600 cm'1
1900 - 2000
strong
4 MASS SPECTROMETRY
It is possible to determine the masses of individual ions in the gas phase. Strictly speaking, it is only possible to measure their mass/charge ratio (m/e), but as multi charged ions are very much less abundant than those with a single electronic charge (e= 1), m/e is for all practical purposes equal to the mass of the ion, m. The principal experimental problems in mass spectrometry are firstly to volatilise the substrate (which implies high vacuum) and secondly to ionise the neutral molecules to charged species.
4.1
IONISATION PROCESSES
The most common method of ionisation involves Electron Impact (El) and there are two general courses of events following a collision of a molecule M with an electron e. By far the most probable event involves electron ejection which yields an odd-electron positively charged cation radical [M]+‘ of the same mass as the initial molecule M. M + e —►
[M]+- + 2e
The cation radical produced is known as the molecular ion and its mass gives a direct measure of the molecular weight of a substance. An alternative, far less probable process, also takes place and it involves the capture of an electron to give a negative anion radical, [M]-. M + e —»
[M]"
Electron impact mass spectrometers are generally set up to detect only positive ions, but negative-ion mass spectrometry is also possible. The energy of the electron responsible for the ionisation process can be varied. It must be sufficient to knock out an electron and this threshold, typically about 10-12 eV, is known as the appearance potential. In practice much higher energies (~70 eV) are used and this large excess energy (1 eV = 95 kJ moh1) causes further fragmentation of the molecular ion. Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
21
Chapter 4 Mass Spectrometry
The two important types of fragmentation are: [M]+‘ —»
A+ (even electron cation) + B' (radical)
[M]+- —>
C+’ (cation radical) + D (neutral molecule)
or
As only species bearing a positive charge will be detected, the mass spectrum will show signals due not only to [M]+- but also due to A+, C+- and to fragment ions resulting from subsequent fragmentation of A+ and C+\ As any species may fragment in a variety of ways, the typical mass spectrum consists of many signals. The mass spectrum consists of a plot of masses of ions against their relative abundance. There are a number of other methods for ionising the sample in a mass spectrometer. The most important alternative ionisation method to electron impact is Chemical Ionisation (Cl). In Cl mass spectrometry, an intermediate substance (generally methane or ammonia) is introduced at a higher concentration than that of the substance being investigated. The carrier gas is ionised by electron impact and the substrate is then ionised by collisions with these ions. Cl is a milder ionisation method than El and leads to less fragmentation of the molecular ion. Another common method of ionisation is Electrospray Ionisation (ESI). In this method, the sample is dissolved in a polar, volatile solvent and pumped through a fine metal nozzle, the tip of which is charged with a high voltage. This produces charged droplets from which the solvent rapidly evaporates to leave naked ions which pass into the mass spectrometer. ESI is also a relatively mild form of ionisation and is very suitable for biological samples which are usually quite soluble in polar solvents but which are relatively difficult to vaporise in the solid state. Electrospray ionisation tends to lead to less fragmentation of the molecular ion than EL Matrix Assisted Laser Desorption Ionisation (M ALDI) uses a pulse of laser light to bring about ionisation. The sample is usually mixed with a highly absorbing compound which acts as a supporting matrix. The laser pulse ionises and vaporises the matrix and the sample to give ions which pass into the mass spectrometer. Again MALDI is a relatively mild form of ionisation which tends to give less fragmentation of the molecular ion than El. All of the subsequent discussion of mass spectrometry is limited to positive-ion electron-impact mass spectrometry.
22
Chapter 4 Mass Spectrometry
4.2
INSTRUMENTATION
In a magnetic sector mass spectrometer (Figure 4.1), the positively charged ions of mass, m, and charge, e (generally e= 1) are subjected to an accelerating voltage V and passed through a magnetic field H which causes them to be deflected into a curved path of radius r. The quantities are connected by the relationship: m
H2r 2
e
2V
The values of FI and V are known, r is determined experimentally and e is assumed to be unity thus permitting us to determine the mass m. In practice the magnetic field is scanned so that streams of ions of different mass pass sequentially to the detecting system (ion collector). The whole system (Figure 4.1) is under high vacuum (less than 10'6 Torr) to permit the volatilisation of the sample and so that the passage of ions is not impeded. The introduction of the sample into the ion chamber at high vacuum requires a complex sample inlet system.
High vacuum
Figure 4.1
Large negative potential
Magnetic Field
Schematic Diagram of an Electron-Impact Mass Spectrometer
23
Chapter 4 Mass Spectrometry
The magnetic scan is synchronised with the x-axis of a recorder and calibrated to appear as mass number (strictly m/e). The amplified current from the ion collector gives the relative abundance of ions on the y-axis. The signals are usually preprocessed by a computer that assigns a relative abundance of 100 % to the strongest peak (base peak).
Many modem mass spectrometers do not use a magnet to bend the ion beam to separate ions but rather use the “time o f flight” (TOF) of an ion over a fixed distance to measure its mass. In these spectrometers, ions are generated (usually using a very short laser pulse) then accelerated in an electric field. Lighter ions have a higher velocity as they leave the accelerating field and their time of flight over a fixed distance will vary depending on the speed that they are travelling. Time of Flight mass spectrometers have the advantage that they do not require large, high-precision magnets to bend and disperse the ion beam so they tend to be much smaller, compact and less complex (desk-top size) instruments.
4.3
MASS SPECTRAL DATA
As well as giving the molecular weight of a substance, the molecular ion of a compound may provide additional information. The "nitrogen rule" states that a molecule with an even molecular weight must contain no nitrogen atoms or an even number of nitrogen atoms. This means that a molecule with an odd molecular weight must contain an odd number of nitrogen atoms. (1)
High resolution mass spectra. The mass of an ion is routinely determined to
the nearest unit value. Thus the mass of [M]+' gives a direct measure of molecular weight. It is not usually possible to assign a molecular formula to a compound on the basis of the integer m/e value of its parent ion. For example, a parent ion at m/e 72 could be due to a compound whose molecular formula is C4H80 or one with a molecular formula C3H40 2 or one with a molecular formula C3H 8N2. However, using a double-focussing mass spectrometer or a time-of-flight mass spectrometer, the mass of an ion or any fragment can be determined to an accuracy of approximately ± 0.00001 of a mass unit (a high resolution mass spectrum). Since the masses of the atoms of each element are known to high accuracy, molecules that may have the same mass when measured only to the nearest integer mass unit, can be distinguished when the mass is measured with high precision. Based on the accurate masses of 12C, 160 , 14N and 'H (Table 4.1) ions with the formulas C4HgO+', C 3H40 2+’ or C3HgN2+’ would have accurate masses 72.0573, 72.0210, and 72.0686 so these
24
Chapter 4 Mass Spectrometry
could easily be distinguished by high resolution mass spectroscopy. In general, if the mass of any fragment in the mass spectrum can be accurately determined, there is usually only one combination of elements which can give rise to that signal since there are only a limited number of elements and their masses are accurately known. By examining a mass spectrum at sufficiently high resolution, one can obtain the exact composition of each ion in a mass spectrum, unambiguously. Most importantly, determining the accurate mass of [M]+' gives the molecular formula o f the compound.
Table 4.1
Isotope
Accurate Masses of Selected Isotopes
Natural
Mass
Abundance (%)
•H
99.98
1.00783
2H 12C
0.016
2.01410
98.9
12.0000
13C
1.1
13.00336
14N
99.6
14.0031
15N 16Q
0.37
15.0001
99.8
15.9949
17Q
0.037
16.9991
18Q 19p
0.20
17.9992
100
18.99840
28Si
92.28
27.9769
29Si
4.7
28.9765
30Si 31P 32$
3.02
29.9738
100
30.97376
95.0
31.9721
33S
0.75
32.9715
34S
4.2
33.9679
35C1
75.8
34.9689
37C1
24.2
36.9659
79Br
50.7
78.9183
81Br
49.3
80.9163
127I
100
126.9045
25
Chapter 4 Mass Spectrometry
(2)
Molecular Fragmentation. The fragmentation pattern is a molecular
fingerprint. In addition to the molecular ion peak, the mass spectrum (see Figure 4.1) consists of a number of peaks at lower mass numbers and these result from fragmentation of the molecular ion. The principles determining the mode of fragmentation are reasonably well understood, and it is possible to derive structural information from the fragmentation pattern in several ways. (a)
The appearance of prominent peaks at certain mass numbers can be correlated empirically with certain structural elements (Table 4.2), e.g. a prominent peak at m/e = 43 is a strong indication of the presence of a CH3-CO- group in the molecule.
(b)
Information can also be obtained from differences between the masses of two peaks. Thus a prominent fragment ion that occurs 15 mass numbers below the molecular ion, suggests strongly the loss of a CH3- group and therefore that a methyl group was present in the substance examined.
(c)
The knowledge of the principles governing the mode of fragmentation of ions makes it possible to confirm the structure assigned to a compound and, quite often, to determine the juxtaposition of structural fragments and to distinguish between isomeric substances. For example, the mass spectrum of benzyl methyl ketone, Ph-CH2-CO-CH3 contains a strong peak at m/e = 91 due to the stable ion Ph-CH2+, but this ion is absent in the mass spectrum of the isomeric propiophenone Ph-CO-CH2CH 3 where the structural elements Ph- and -CH2- are separated. Instead, a prominent peak occurs at m/e = 105 due to the stable ion Ph-C=0+.
Electronic databases of the mass spectral fragmentation patterns of known molecules can be rapidly searched by computer. The pattern and intensity of fragments in the mass spectrum is characteristic of an individual compound so comparison of the experimental mass spectrum of a compound with those in a library can be used to positively identify it, if its spectrum has been recorded previously.
26
Chapter 4 Mass Spectrometry
Table 4.2
Common Fragments and their Masses
Fragment
Mass
Fragment
Mass
Fragment
Mass
O II
CH3-
15
CH3CH2-
29
NO
30
— CH2OH
31
/C —
29
CH2=CH-CH2
41
H
O II
o II
43
.c — ch
HO
—N 02
45
46
3^ o
C 4h7
.OH
55
C 4H9
57
60
C 5H5
65
CH3CH2^
/
_
\ _
57
77
ch 2= c
X° H
c 6h 5
_
91
xx - f u
N O ^ c h 2-
1%) of minor isotopes, is often obvious simply by inspection of ions near the molecular ion.
27
Chapter 4 Mass Spectrometry
The relative intensities of the [M]+‘, [M+l]+‘ and [M+2]+‘ ions exhibit a characteristic pattern depending on the specific isotopes that make up the ion. For any molecular ion (or fragment) which contains one bromine atom, the mass spectrum will contain two peaks separated by two m/e units, one for the ions which contain 79Br and one for the ions which contain 81Br. For bromine-containing fragments, the relative intensities of the two ions will be approximately the same, since the natural abundances of 79Br and 8lBr are 100
approximately equal. In the mass spectrum of 1-bromoethane, there are two molecular ions of almost equal intensity. The peak at m/e 108 corresponds to the molecular ions in the sample which contain 79Br; the peak at
Mass Spectrum
60 40
C 2H 581B r+ '
CO - 0 - CL - (D ■ 0 - CO . -Q 7 o “ as
M = 108 C 2H 579Br+ '
20 7 u li ____ ___i__i__i__i__i__i__i__i__i__.__i---------------
40
m/e 110 corresponds to the molecular ions in the sample which contain
81
M = 110
C 2H 5Br
80 -•
80 m /e
120
Br.
Similarly, for any molecule (or fragment of an molecule) which contains one chlorine atom, the mass spectrum will contain two fragments separated by two m/e units, one for the ions which contain 35Cl and one for the ions which contain 37C1. For chlorinecontaining ions, the relative intensities of the two ions will be approximately 3:1 since this reflects the natural abundances of 35C1 and 37C1. In the mass spectrum of 2-chloropropane, there are two molecular ions at 78 and 80
100 -
Mass Spectrum 43
with intensities approximately in the ratio
80 -•
3:1. The peak at m/e 78 corresponds to the
60 »
molecular ions in the sample which contain
40
35C1; the peak at m/e 80 corresponds to the
20
CO
- 0 . CL - 0 .
M = 78
0 CO
c
-Q 7 o
M = 80 c ^ 3, nh
molecular ions in the sample which contain 37C1. Note that the base peak at m/e 43 is
only a single ion so this ion must contain no
35^.+' ci
3h / °
40
^ f 7737c i + ‘ I I I—a L 120 80
m /e
chlorine. The pair of ions at m/e 63 and 65 clearly corresponds to a fragment that still contains a chlorine atom. Any molecular ion (or fragment) which contains 2 bromine atoms will have a pattern of ions M:M+2:M+4 with signals in the ratio 1:2:1 and any molecular ion (or fragment) which contains 2 chlorine atoms will have a pattern of M:M+2:M+4 with signals in the ratio 10:6.5:1 (Figure 4.2).
28
Chapter 4 Mass Spectrometry
Br
Brc CM +
O MT O CD
Figure 4.2
+
T- oo ^ r to
Cl
B ra CM NT CD + + +
CO O
CD CO
Cl2
CM
+
O o
vCO
O LOo O
CD t —
Cl3
BrCI
CM NT CD + + +
CM
O LO-r- CO OCJ) CO
OOOMN O CM
+ +
BrCI2 CM
CD
+++
Br2CI CM
CD
+++
CM O L O CD CD O M -
Relative Intensities of the Cluster of Molecular Ions for Molecules Containing Combinations of Bromine and Chlorine Atoms
(4)
Chromatography coupled with Mass Spectrometry. It is now common to
couple an instrument for separating a mixture of organic compounds e.g. using gas chromatography (GC) or high performance liquid chromatography (HPLC), directly to the input of a mass spectrometer. In this way, as each individual compound is separated from the mixture, its mass spectrum can be recorded and compared automatically with the library of known compounds and identified immediately if it is a known compound. (5)
Metastable peaks in a mass spectrum arise if the fragmentation process a+ —> b+ + c (neutral)
takes place within the ion-accelerating region of the mass spectrometer (Figure 4.1). Ion peaks corresponding to the masses of a+ and to b+ (ma and /??/,) may be accompanied by a broader peak at mass m , such that: 2
mb m* = ------ma
The presence of metastable peaks in a mass spectrum often permits positive identification of a particular fragmentation path.
4.4
REPRESENTATION OF FRAGMENTATION PROCESSES
As fragmentation reactions in a mass spectrometer involve the breaking of bonds, they can be represented by the standard "arrow notation" used in organic chemistry. For some purposes a radical cation (e.g. a generalised ion of the molecular ion) can be represented without attempting to localise the missing electron: [M]+- or
[H3C-CH2-0-R] +•
29
Chapter 4 Mass Spectrometry
However, to show a fragmentation process it is generally necessary to indicate "from where the electron is missing" even though no information about this exists. In the case of the molecular ion corresponding to an alkyl ethyl ether, it can be reasonably inferred that the missing electron resided on the oxygen. The application of standard arrow notation permits us to represent a commonly observed process, viz. the loss of a methyl fragment from the [H3C-CH2-0 -R ]+' molecular ion:
4.5
FACTORS GOVERNING FRAGMENTATION PROCESSES
Three factors dominate the fragmentation processes: (a)
Weak bonds tend to be broken most easily
(b)
Stable fragments (not only ions, but also the accompanying radicals and molecules) tend to be formed most readily
(c)
Some fragmentation processes depend on the ability of molecules to assume cyclic transition states.
Favourable fragmentation processes naturally occur more often and ions thus formed give rise to strong peaks in the mass spectrum.
4.6
EXAMPLES OF COMMON TYPES OF FRAGMENTATION
There are a number of common types of cleavage which are characteristic of various classes of organic compounds. These result in the loss of well-defined fragments which are characteristic of certain functional groups or structural elements. (1)
Cleavage at Branch Points. Cleavage of aliphatic carbon skeletons at branch
points is favoured as it leads to more substituted (and hence more stable) carbocations. The mass spectrum of 2,2-dimethylpentane shows strong peaks at m/e = 85 and m/e = 57 where cleavage leads to the formation of stable tertiary carbocations.
30
Chapter 4 Mass Spectrometry
CH3
ch3
I
+• I
CH3— C—CHo—CHo—c h 3
c h 3—c —c h 2—c h 2—c h 3
cm
ch3
m/e = 100
ch3
CH, ch3
c h 3—c —c h 2- c h 2- c h 3 ch3
C—CH2—CH2—CH3 ch3
neutral fragment
stable cation
m/e = 85 ch3 I+
CH3 +• I c h 3—c —c h 2—c h 2—c h 3
cm—c
CH
ch
CH,
stable cation
2— c h 2— c h 3
neutral fragment
m/e = 57 - Cleavage. Chain cleavage tends to occur P to heteroatoms, double bonds and aromatic rings because relatively stable, delocalised carbocations result in each case. (a) .. I I R - X — C — C— I I
X=
R - X — C— CI I
0 , N, S, halogen + / R -X =C n
+ / R -X -C n
\
resonance stabilised carbocation
(b)
V - r / / C- ° \
neutral fragment
' + I
c— c-
x l
-e-
/
.C-C
I
W
\ /
/
C— C \
. +
s +
c—
« t C -C *— ► /
/
I
l
c— I
resonance stabilised carbocation
. / C—
\
neutral fragment
Chapter 4 Mass Spectrometry
(c)
\C — C I — I
a
- e -
I
resonance stabilised carbocation
(3)
Cleavage a to carbonyl groups. Cleavage tends to occur a to carbonyl
groups to give stable acylium cations. R may be an alkyl, -OH or -OR group.
.C
^
^
R /
c C \
^
c R
I ^ a
resonance stabilised carbocation
/
*c— \ neutral fragment
O II C + I R
(4)
Cleavage a to heteroatoms. Cleavage of chains may also occur a to
heteroatoms, e.g. in the case of ethers:
I R -O -C " I
(5)
. R -O -C " |
► R -0: C" I free radical carbocation
Retro Diels-Alder reaction. Cyclohexene derivatives may undergo a retro
Diels-Alder reaction:
32
-e'
Chapter 4 Mass Spectrometry
(6)
The McLafferty rearrangement. Compounds where the molecular ion can
assume the appropriate 6 -membered cyclic transition state usually undergo a cyclic fragmentation, known as the McLafferty rearrangement. This rearrangement involves a transfer of a y hydrogen atom to an oxygen and is often observed with ketones, acids and esters:
With primary carboxylic acids, R-CHr COOH, this fragmentation leads to a characteristic peak at m/e = 60 +
h2c = c — oh
OH With carboxylic esters, two types of McLafferty rearrangements may be observed and ions resulting from either fragmentation pathway are commonly observed in the mass spectrum:
5 NUCLEAR MAGNETIC RESONANCE (NMR) SPECTROSCOPY
5.1
THE PHYSICS OF NUCLEAR SPINS AND NMR INSTRUMENTS
(1)
The Larmor Equation and Nuclear Magnetic Resonance
All nuclei have charge because they contain protons and some of them also behave as if they spin. A spinning charge generates a magnetic dipole and is associated with a small magnetic field H (Figure 5.1). Such nuclear magnetic dipoles are characterised by nuclear magnetic spin quantum numbers which are designated by the letter I and can take up values equal to 0 , V2, 1 , 3/2 ... etc.
H
Figure 5.1
A Spinning Positive Charge Generates a Magnetic Field and Behaves like a Small Magnet
It is useful to consider three types of nuclei: Type 1:
Nuclei with 1 = 0. These nuclei do not interact with the applied magnetic field and are not NMR chromophores. Nuclei with 1 = 0 have an even number of protons and even number of neutrons and have no net spin. This means that nuclear spin is a property characteristic of certain isotopes rather than of certain elements. The most prominent examples of nuclei with I = 0 are !2C and 160 , the dominant isotopes of carbon and oxygen. Both oxygen and carbon also have isotopes that can be observed by NMR spectroscopy.
34
Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
Chapter 5 NMR Spectroscopy
Type 2:
Nuclei with I = V2. These nuclei have a non-zero magnetic moment and are NMR visible and have no nuclear electric quadrupole (Q). The two most important nuclei for NMR spectroscopy belong to this category: *H (ordinary hydrogen) and 13C (a non-radioactive isotope of carbon occurring to the extent of 1.06% at natural abundance). Also, two other commonly observed nuclei 19F and 31P have I = A Together, NMR data for 'H and 13C account for well over 90% of all NMR observations in the literature and the discussion and examples in this book mostly refer to these two nuclei. However, the spectra of all nuclei with I = x/i can be understood easily on the basis of common theory.
Type 3:
Nuclei with I > ‘A These nuclei have both a magnetic moment and an electric quadrupole. This group includes some common isotopes (e.g. 2H and 14N) but they are more difficult to observe and spectra are generally very broad. This group of nuclei will not be discussed further.
The most important consequence of nuclear spin is that in a uniform magnetic field, a nucleus of spin I may assume 21 + 1 orientations. For nuclei with I = A there are just 2 permissible orientations (since 2 x y2 + 1 = 2). These two orientations will be of unequal energy (by analogy with the parallel and antiparallel orientations of a bar magnet in a magnetic field) and it is possible to induce a spectroscopic transition (spin-flip) by the absorption of a quantum of electromagnetic energy (AE) of the appropriate frequency (v):
h
(5-1)
In the case of NMR, the energy required to induce the nuclear spin flip also depends on the strength of the applied field, H 0. It is found that: v = KHq
(5.2)
where ATis a constant characteristic of the nucleus observed. Equation 5.2 is known as the Larmor equation and is the fundamental relationship in NMR spectroscopy. Unlike other forms of spectroscopy, in NMR the frequency of the absorbed electromagnetic radiation is not anabsolute value for anyparticular transition, but has a different value depending on thestrength of the applied magnetic field. For every value of H 0, there is a matching value of v corresponding to the condition of resonance according to Equation 5.2, and this is the origin of the term “resonance ” in Nuclear Magnetic Resonance Spectroscopy. Thus for 'H and 13C, resonance
35
Chapter 5 NMR Spectroscopy
frequencies corresponding to magnitudes of applied magnetic field (H 0) commonly found in commercial instruments are given in Table 5.1.
Table 5.1
Resonance Frequencies of *H and 13C Nuclei in Magnetic Fields of Different Strengths
v *H (MHz)
v 13C (MHz)
H 0 (Tesla)
60
15.087
1.4093
90
22.629
2.1139
100
25.144
2.3488
200
50.288
4.6975
300
75.432
7.0462
400
100.577
9.3950
500
125.720
11.744
600
150.864
14.0923
750
188.580
17.616
800
201.154
18.790
900
226.296
21.128
In common jargon, NMR spectrometers are commonly known by the frequency they use to observe ’H i.e. as "60 MHz", "200 MHz" or "400 MHz" instruments, even if the spectrometer is set to observe a nucleus other than 'H. All the frequencies listed in Table 5.1 correspond to the radio frequency region of the electromagnetic spectrum and inserting these values into Equation 5.1 gives the size of the energy gap between the states in an NMR experiment. A resonance frequency of 100 MHz corresponds to an energy gap of approximately 4 x 10’5 kJ mob1. This is an extremely small value on the chemical energy scale and this means that NMR spectroscopy is, for all practical purposes, a ground-state phenomenon. Any absorption signal observed in a spectroscopic experiment must originate from excess of the population in the lower energy state, the so called Boltzmann excess, which is equal to N^-Na, where iVp and Na are the populations in the lower ((3) and upper (a) energy states.
36
Chapter 5 NMR Spectroscopy
For molar quantities, the general Boltzmann relation (Equation 5.3) shows that: AE iV p
RT
=
e
(5 J )
N H —
area = 3
1 : 2:1
-CHdoublet area = 3
1 :3 :3 : 1
1:1
—CH2—
-C H 2-
triplet
triplet
area = 2
area = 2
1 : 2:1
1 : 2:1
-(CH3)2
/CH3
doublet
—CH
area = 6
xCH3
an isopropyl grc
1 : 6 : 15 : 20 : 15 : 6 : 1
1:1
55
Chapter 5 NMR Spectroscopy
5.7
ANALYSIS OF 1H NMR SPECTRA
To obtain structurally useful information from NMR spectra, one must solve two separate problems. Firstly, the spectrum must be analysed to obtain the NMR parameters (chemical shifts and coupling constants) for all the protons and, secondly, the values of the coupling constants must be interpreted in terms of established relationships between the parameters and structure. (1)
A spin system is defined as a group of coupled protons. Clearly, a spin system
cannot extend beyond the bounds of a molecule, but it may not include a whole molecule. For example, isopropyl propionate comprises two separate and isolated proton spin systems, a seven-proton system for the isopropyl residue and a five-proton system for the propionate residue, because the ester group effectively provides a barrier (5 bonds) against coupling between the two parts.
c h 3x CH O -C CH3x 7-spin system
(2)
ii
CH2CH3 Isopropyl propionate
O 5-spin system
Strongly and weakly coupled spins. These terms refer not to the actual
magnitude of J, but to the ratio of the separation of chemical shifts expressed in Hz (Av) to the coupling constant J between them. For most purposes, if Av/J is larger than ~3, the spin system is termed weakly coupled. When this ratio is smaller than ~3, the spins are termed strongly coupled. Two important conclusions follow: (a)
Because the chemical shift separation (Av) is expressed in Hz, rather than in the dimensionless 8 units, its value will change with the operating frequency of the spectrometer, whiie the value of J remains constant. It follows that two spins will become progressively more weakly coupled as the spectrometer frequency increases. Weakly coupled spin systems are much easier to analyse than strongly coupled spin systems and thus spectrometers operating at higher frequencies (and therefore at higher applied magnetic fields) will yield spectra which are more easily interpreted. This has been an important reason for the development of NMR spectrometers operating at ever higher magnetic fields.
56
Chapter 5 NMR Spectroscopy
(b)
Within a spin system, some pairs of nuclei or groups of nuclei may be strongly coupled and others weakly coupled. Thus a spin-system may be partially strongly coupled.
(3)
Magnetic equivalence. A group of protons is magnetically equivalent when
they not only have the same chemical shift (chemical equivalence) but also have identical spin-spin coupling to each individual nucleus outside the group. (4)
Conventions used in naming spin systems. Consecutive letters of the
alphabet (e.g. A, B, C D ,
) are used to describe groups of protons which are
strongly coupled. Subscripts are used to give the number of protons that are magnetically equivalent. Primes are used to denote protons that are chemically equivalent but not magnetically equivalent. A break in the alphabet indicates weakly coupled groups. For example: ABC denotes a strongly coupled 3-spin system AMX denotes a weakly coupled 3-spin system ABX denotes a partially strongly coupled 3-spin system A3BMXY denotes a spin system in which the three magnetically equivalent A nuclei are strongly coupled to the B nucleus, but weakly coupled to the M, X and Y nuclei. The nucleus X is strongly coupled to the nucleus Y but weakly coupled to all the other nuclei. The nucleus M is weakly coupled to all the other 6 nuclei. AA'XX' is a 4-spin system described by two chemical shift parameters (for the nuclei A and X) but where JAX ^ JAX,. A and A' (as well as X and X') are pairs of nuclei which are chemically equivalent but magnetically non-equivalent. The process of deriving the NMR parameters (8 and J) from a set of multiplets in a spin system is known as the analysis o f the NMR spectrum. In principle, any spectrum arising from a spin system, however complicated, can be analysed but some will require calculations or simulations performed by a computer. Fortunately, in a very large number of cases, multiplets can be correctly analysed by inspection and direct measurements. These spectra are known as first order spectra and they arise from weakly coupled spin systems. At high applied magnetic fields, a large proportion of 'H NMR spectra are nearly pure first-order and there is a tendency for simple molecules, e.g. those exemplified in the problems in this text, to exhibit first-order spectra even at moderate fields.
57
Chapter 5 NMR Spectroscopy
5.8
CHANGING THE MAGNETIC FIELD IN NMR SPECTROSCOPY
Changing the magnetic field of an NMR spectrometer i.e. recording the NMR spectrum of a sample on NMR instruments operating at different magnetic fields, highlights several important aspects of NMR spectroscopy. The chemical shifts of the nuclei (expressed in Hz) are proportional to the strength of the magnetic field (see Section 5.1). The higher the magnetic field, the higher the resonance frequency of each nucleus in the sample. Nuclei move further apart (in Hz) from each other as the magnetic field strength increases i.e. there is better dispersion in the spectrum. On the other hand, spin-spin coupling is a molecular property that is independent of the magnetic field of the spectrometer. So the splittings in an NMR spectrum due to spin-spin coupling, remain constant, irrespective of the magnetic field strength of the spectrometer. Given below are H NMR spectra of 2-bromotoluene where spectra on the same sample have been recorded in 3 different NMR spectrometers each operating at a different magnetic field strength. There are 4 different aromatic proton resonances. At the lowest field (4.7 Tesla), there is significant overlap of the resonances. The resonances for H3 and H 6 are severely distorted and this is not a 1st order spectrum. At 14.1 Tesla (600 MHz) dispersion is much better and the spectrum is a 1st order spectrum which could be analysed by 1st order rules. 600 MHz spectrometer 14.1 Tesla magnet
Aromatic region of the 'H NMR spectrum of 2 -bromotoluene
400 MHz spectrometer 9.4 Tesla magnet
(acetone-^, solution).
200 MHz spectrometer 4.7 Tesla magnet
Chapter 5 NMR Spectroscopy
5.9
Rule 1
RULES FOR SPECTRAL ANALYSIS OF FIRST ORDER SPECTRA
A group of n magnetically equivalent protons will split a resonance of an interacting group of protons into n + 1 lines. For example, the resonance due to the A protons in an A nX m system will be split into m+ 1 lines, while the resonance due to the X protons will be split into n+ 1 lines. More generally, splitting by n nuclei of spin quantum number I, results in 2 «I +1 lines. This simply reduces to n+1 for protons where I = V2.
Rule 2
The spacing (measured in Hz) of the lines in the multiplet will be equal to the coupling constant. In the above example, all spacings in both parts of the spectrum will be equal to JAX.
Rule 3
The true chemical shift of each group of interacting protons lies in the centre of the (always symmetrical) multiplet.
Rule 4
The relative intensities of the lines within each multiplet will be in the ratio of the binomial coefficients (Table 5.9). Note that, in the case of higher multiplets, the outside components of multiplets are relatively weak and may be lost in the instrumental noise, e.g. a septet may appear as a quintet if the outer lines are not clearly visible. The intensity relationship is the first to be significantly distorted in non-ideal cases, but this does not lead to serious errors in spectral analysis.
Rule 5
When a group of magnetically equivalent protons interacts with more than one group of protons, its resonance will take the form of a multiplet o f multiplets. For example, the resonance due to the A protons in a system AW M XOTwill have the multiplicity of (p+X){m+ \). The multiplet patterns are chained e.g. a proton coupled to 2 different protons will be split to a doublet by coupling to the first proton then each of the component of the doublet will be split further by coupling to the second proton resulting in a symmetrical multiplet with 4 lines (a doublet of doublets).
59
Chapter 5 NMR Spectroscopy
Ha U
LI
yM | yx X -C -C -C -Y doublet of doublets
Ha |_| J yM | L yx X -C -C -C -Y HM 1 1
triplet of doublets or doublet of triplets
The appropriate coupling constants will control splitting and relative intensities will obey rule 4. Rule
6
Rule 7
Protons that are magnetically equivalent do not split each other. Any system A„ will give rise to a singlet. Spin systems that contain groups of chemically equivalent protons that are not magnetically equivalent cannot be analysed by first-order methods.
Rule
8
If Avab/Tab is less than ~3, for any pair of nuclei A and B in the spin system, the spectra become distorted from the expected ideal multiplet patterns and the spectra cannot be analysed by first-order methods.
60
Chapter 5 NMR Spectroscopy
(1) S p littin g D ia g ra m s
The knowledge of the rules listed above, permits the development of a simple procedure for the analysis of any spectrum which is suspected of being first order. The first step consists of drawing a splitting diagram, from which the line spacings can be measured and identical (hence related) splittings can be identified (Figure 5.8).
8 (ppm from TMS)
Figure 5.8
A Portion of the *H NMR Spectrum of Styrene Epoxide (100 MHz as a 5% solution in CCI4)
The section of the spectrum of styrene epoxide (Figure 5.8) clearly contains the signals from 3 separate protons (identified as Hi, H 2 and H3) with Hi at 5 2.95, H2 at 5 2.58 and H 3 at 5 3.67 ppm. Each signal appears as a doublet of doublets and the chemical shift of each proton is simply obtained by locating the centre of each multiplet. The pair of nuclei giving rise to each splitting is clearly indicated by the splitting diagram above each multiplet with 2J H,_H2 = 5.9 Hz, V H1.H3 = 4.0 Hz and = 2.5 Hz. The validity of a first order analysis can be verified by calculating the ratio Av/J for each pair of nuclei and establishing that it is greater than 3 .
61
Chapter 5 NMR Spectroscopy
From Figure 5.8 ^
.
IL
.
5.9
J-\ 2
6.3
^13 =
72
J-13
4.0
= ig o
Av23 _ 109 = 43 6
J 23 2.5
Each ratio is greater than 3 soa first order analysis is justified and the 100 MHz spectrum of the aliphatic protons of styrene oxide is indeeda first order spectrum and could be labelled as an AMX spin system. The 60 MHz ’H spectrum of a 4 spin AMX2 system is given in Figure 5.9. This system contains 3 separate proton signals (in the intensity ratios 1 : 1 : 2 , identified as Ha, Hm and Hx). The multiplicity of HA is a triplet of doublets, the multiplicity of HM is a triplet of doublets and the multiplicity of Hx is a doublet of doublets. Again, the nuclei giving rise to each splitting are clearly indicated by the splitting diagram above each multiplet and the chemical shifts of each multiplet are simply obtained by measuring the centres of each multiplet.
Hx 8 = 7.0 ppm
8 = 6.0 ppm
8 = 4.95 ppm
v = 420 Hz
v = 360 Hz
v = 297 Hz
7.0
Figure 5.9
62
6.0
5.0
8 (ppm from TMS)
The 60 MHz 'H NMR Spectrum of a 4-Spin AMX2 Spin System
Chapter 5 NMR Spectroscopy
A spin system comprising just two protons (i.e. an AX or an AB system) is always exceptionally easy to analyse because, independent of the value of the ratio of Av/J, the spectrum always consists of just four lines with each pair of lines separated by the coupling constant J. The only distortion from the first-order pattern consists of the gradual reduction of intensities of the outer lines in favour of the inner lines, a characteristic "sloping" or "tenting" towards the coupling partner. A series of simulated spectra of two-spin systems are shown in Figure 5.10.
12
Av12
—
J
Ul
= 10.0
J12
Av.12
3.0
12
uu Av12 — 12 = 0.8 J12 jU ul
Av12 — 2 = 0.0 J 12
-j----- 1------- 1----1-------1---- 1----- 1----- 1------1— 100
Figure 5.10
50
Simulated
0 Hz
-50
-100
NMR Spectra of a 2-Spin System as the Ratio
Av/J, is Varied from 10.0 to 0.0
63
Chapter 5 NMR Spectroscopy
C oin cid en ta lly e q u a l va lu es o f c o u p lin g constants.
An AMX spin system. First order analysis rules predict that the resonance for Hm in an AMX spin system will be a doublet of doublets (4 lines) since HMwill be split by coupling to Ha and to Hx- All lines of the multiplet will have equal intensity and the spacings in the multiplet will be J a m and J ax However, in the case where J am and J ax are approximately equal, the central lines of the multiplet overlap to give a single line whose intensity is twice as high as the outer lines.
While the multiplet is
A doublet of doublets appears like a triplet if
technically a doublet of
approximately equal
Jamand ^mxare
doublets, it appears as a triplet
Hm
(3 lines) with intensities in the ratio 1 :2 : 1 .
An A 2M 2X2 spin system. First order analysis rules predict that the resonance for Hm in an A 2M 2X2 spin system will be a triplet of triplets (9 lines) since Hm will be split by coupling to 2 x Ha nuclei and to 2 x Hx nuclei. The relative intensities of the lines in the multiplet can be predicted easily using Table 5.9. The spacings in the multiplet will be equal to J a m and J a x However, in the case where Jam and J ax are approximately equal, there is overlap between the lines of the multiplet. A triplet of triplets appears like a quintet if Jam and JMXare approximately equal
While the multiplet is technically a triplet of triplets, it appears as a
H*
quintet (5 lines) with 4I
intensities in the ratio 1:4:6:4 1.
1 2 Ll 1 2
1\
1 I
This is not an uncommon situation in flexible alkyl chains (X-CH2-CH2-CH 2-Y) since the 3 -bond vicinal coupling between protons on adjacent carbons typically falls within a narrow range of about 6-8 Hz.
64
Chapter 5 NMR Spectroscopy
(2) Spin Decoupling
In the signal of a proton that is a multiplet due to spin-spin coupling, it is possible to remove the splitting effects by irradiating the sample with an additional Rf source at the exact resonance frequency of the proton giving rise to the splitting. The additional radiofrequency causes rapid flipping of the irradiated nuclei and as a consequence nuclei coupled to them cannot sense them as being in either an a or p state for long enough to cause splitting. The irradiated nuclei are said to be decoupled from other nuclei in the spin system. Decoupling simplifies the appearance of complex multiplets by removing some of the splittings. In addition, decoupling is a powerful tool for assigning spectra because the skilled spectroscopist can use a series of decoupling experiments to sequentially identify which nuclei are coupled. In a 4-spin AM2X spin system, the signal for proton HAwould appear as a doublet of triplets (with the triplet splitting due to coupling to the 2 M protons and the doublet splitting due to coupling to the X proton). Irradiation at the frequency of Hx reduces the multiplicity of the A signal to a triplet (with the remaining splitting due to ./AM) and irradiation at the frequency of HMreduces the multiplicity of the A signal to a doublet (with the remaining splitting due to JAX) (Figure 5.11). Ha
Basic spectrum of Ha without irradiation of of H m or Hx
Figure 5.11
with irradiation of Hx
with irradiation of HM
Selective Decoupling in a Simple 4-Spin System
65
Chapter 5 NMR Spectroscopy
5.10
CORRELATION OF 1H - 1H COUPLING CONSTANTS WITH STRUCTURE
Interproton spin-spin coupling constants are of obvious value in obtaining structural data about a molecule, in particular information about the connectivity of structural elements and the relative disposition of various protons.
Non-aromatic Spin Systems. In saturated systems, the magnitude of the geminal coupling constant -VH_C_H(two protons attached to the same carbon atom) is typically between 10 and 16 Hz but values between 0 and 22 Hz have been recorded in some unusual structures.
FL
Ha C 'C
FT' n H b
2
Jab =10-16 Hz.
The vicinal coupling (protons on adjacent carbon atoms) V H.C.C.Hcan have values 0 - 16 Hz depending mainly on the dihedral angle for 90 < ()>< 180°
IN m8 < ^ 4 o
0
45 90 135 180 Dihedral angle(tj)) (degrees)
It follows from these equations that if the dihedral angle between two vicinal protons is near 90° then the coupling constant will be very small and conversely, if the dihedral angle Ai-c=c-H(tram) not overlap. This means that the stereochemistry of the double bond can be determined by measuring the coupling constant between vinylic protons. Where the C=C bond is in a ring, the 5J H.C=C.Hcoupling reflects the ring size. Ha
H \
J,'AB(cis)
/ Hb / C=C\ X Hc
%AC(trans) = 12-19 Hz J\BC(gem) = 0 - 3 Hz
= 5 - 7 Hz
JAB(cis)
= 6 - 11 Hz Hb Ha
I
=9-11
JAB(cis)
HZ
Hb
In alkyl-substituted alkenes, the long-range allylic couplings, (4J ab and 4J ac ) are typically in the range 0-3 Hz. In systems which are stereochemically constrained, the magnitude of the coupling is a function of the dihedral angle between the C-HAbond and the plane of the double bond in a
\
y ha C= C 7
\
He
4j AB] 4j AC = 0 - 3 Hz
relationship reminiscent of the Karplus relation.
67
Chapter 5 NMR Spectroscopy
Aromatic Spin Systems The coupling constant between protons attached to an aromatic ring is diagnostic of the relative position of the coupled protons i.e. whether they are ortho, meta or para. Ha
Ha
J,AB(meta)
= 1 - 3 Hz
AB(para)
0 -1 .5 Hz
Similarly in condensed polynuclear aromatic compounds and heterocyclic compounds, the magnitude of the coupling constants between protons in the aromatic rings reflects the relative position of the coupled protons.
SJ 1,2 = 8 .3 - 9 .1 Hz
3J2 ,3 = 4.0 - 5.7 Hz
3J2i3 = 4.7 Hz
3J2>3 = 1.8 Hz
3J2 ,3 = 6.1 - 6 . 9 Hz
3J3,4 = 6.8 - 9.1 Hz
3J3,4 = 3.4 Hz
3J3,4 = 3.5 Hz
4J1i3= 1 .2 - 1 .6 Hz
4J2 ,4 = 0.0 - 2.5 Hz
4J2,4 = 1.0 Hz
4J2,4 = 0.8 Hz
5J1i4 = 0 - 1 . 0 Hz
4J3,5 = 0 .5 - 1 . 8 Hz
4J2,5 = 2.9 Hz
4J2,5 = 1.6 Hz
5J1iS = 0 - 1 . 5 Hz
4J2,6 = 0.0 - 0.6 Hz
4J2,4; 4d2,6 = 0 Hz
d2i3 = 4.3 Hz
5J2 ,5 = 1.5 Hz
4J2 ,4 = 1.8 Hz
3J4 si 3J s,g = 5.0 Hz
3J3i4 = 8.3 Hz
5J2 ,5 = 0.0 - 2.3 Hz 4
n
h- n^
\
n
3
r\
M
5^
N H
4Jia H and 13C nuclei in a sample. The resulting 13C DEPT spectrum contains only signals arising from protonated carbons (non protonated carbons do not give signals in the 13C DEPT spectmm). The signals arising from carbons in CH 3 and CH groups (i.e. those with an odd number of attached protons) appear oppositely phased from those in CH 2 groups (i.e. those with an even number of attached protons) so signals from CH 3 and CH groups point upwards while signals from CH2 groups point downwards (Figure 6.1b). In more advanced applications, the 13C DEPT experiment can be used to separate the signals arising from carbons in CH3, CH2 and CH groups. This is termed spectral
72
Chapter 6 13C NMR Spectroscopy
editing and can be used to produce separate 13C sub-spectra of just the CH3 carbons, just the CH2 carbons or just the CH carbons. Figure 6.1 shows various 13C spectra of methyl cyclopropyl ketone. The 13C spectrum acquired with full proton decoupling (Figure 6.1a) shows 4 singlet peaks, one for each of the 4 different carbon environments in the molecule. The DEPT spectrum (Figure 6.1b) shows only the 3 resonances for the protonated carbons. The carbon atoms that have an odd number of attached hydrogens (CH and CH3 groups) point upwards and those with an even number of attached hydrogen atoms (the signals of CH2 groups) point downwards. Note that the carbonyl carbon does not appear in the DEPT spectrum since it has no attached protons. In the carbon spectrum with no proton decoupling (Figure 6.1c), all of the resonances of protonated carbons appear as multiplets and the multiplet structure is due to coupling to the attached protons. The CH3 (methyl) group appears as a quartet, the CH2 (methylene) groups appear as a triplet and the CH (methine) group appears as a doublet while the carbonyl carbon (with no attached protons) appears as a singlet. In Figure 6.1c, all of the !JC_Hcoupling constants could be measured directly from the spectrum. CH, (c) with 1H fully coupled
o II c
CH CH c h 2^
;c=o
- ch /
x ch3
ch2
(b)
DEPT
ch2(,
CH3f CHf
(a) with 1H fully decoupled
I 215
205
Figure 6.1
30
T— i 20
10
r ppm
13C NMR Spectra of Methyl Cyclopropyl Ketone (CDCI3 Solvent, 100 MHz), (a) with Broad Band Decoupling of *H ; (b) DEPT Spectrum (c) with no Decoupling of 1H.
73
Chapter 6 13C NMR Spectroscopy
For purposes of assigning a 13C spectrum, two 13C spectra are usually obtained. Firstly, a spectrum with complete *H decoupling to maximise the intensity of signals and provide sharp singlets to minimise any signal overlap. This is the best spectrum to count the number of resonances and accurately determine their chemical shifts. Secondly, a spectrum which is sensitive to the number of protons attached to each C to permit partial sorting of the 13C signals according to whether they are methyl, methylene, methine or quaternary carbon atoms. This could be a DEPT spectrum or a 13C spectrum with no proton decoupling. The number of resonances visible in a 13C NMR spectrum immediately indicates the number of distinct 13C environments in the molecule (Table 6.1). If the number of ,3C environments is less than the number of carbons in the molecule, then the molecule must have some symmetry that dictates that some 13C nuclei are in identical environments. This is particularly useful in establishing the substitution pattern (position where substituents are attached) in aromatic compounds.
Table 6.1
The Number of Aromatic 13C Resonances in Benzenes with Different Substitution Patterns
Molecule
o o-° a:
Number of aromatic ,3C resonances
1
Cl—
4
Br^
\
74
Cl
Number of aromatic 13C resonances
—Cl
2
a
4
y
Cl 3
/Cl
0
Molecule
0
6
Br Cl
4
Br
6
Chapter 6 13C NMR Spectroscopy
6.3
SHIELDING AND CHARACTERISTIC CHEMICAL SHIFTS IN 13C NMR SPECTRA
The general trends of l3C chemical shifts somewhat parallel those in 'H NMR spectra. However, l3C nuclei have access to a greater variety of hybridisation states (bonding geometries and electron distributions) than 'H nuclei and both hybridisation and changes in electron density have a significantly larger effect on l3C nuclei than 'H nuclei. As a consequence, the l3C chemical shift scale spans some 250 ppm, cf. the 10 ppm range commonly encountered for 'H chemical shifts (Tables 6.2 - 6.4, Figure 6.2). Table 6.2
Typical l3C Chemical Shift Values in Selected Organic Compounds Compound
8 ,3C (ppm from TMS)
CH4 CH3CH3 CH3OH CH 3CI CH2C12 CHCl,
-2.1 7.3 50.2 25.6 52.9 77.3
CH 3CH2CH2C1
11.5 (CH3) 26.5 (-CH2-) 46.7 (-CH 2-C1)
c h 2= c h 2
123.3
c h 2= c = c h 2
208.5 (=C=) 73.9 (=CH2)
CH3CHO
31.2 (-CH3) 200.5 (-CHO)
CH 3COOH
20.6 (-CH3), 178.1 (-COOH)
CH3COCH3
30.7 (-CH3), 206.7 (-CO-) 128.5
0 3 4 0\ = /
2
149.8 (C-2) 123.7 (C-3) 135.9 (C4)
75
Chapter 6 13C NMR Spectroscopy
Table 6.3
Typical ,3C Chemical Shift Ranges in Organic Compounds
Group
13C shift (ppm)
TMS
0.0
-CH 3 (with only -H or -R at Ca and Cp)
0-30
-CH2 (with only -H or -R at Ca and Cp)
20-45
-CH (with only -H or -R at Ca and Cp)
30-60
C quaternary (with only -H or -R at Ca and Cp)
30-50
o -c h 3
50-60
n -ch 3
15-45
oc
70 - 95
c=c
105 - 160
C (aromatic)
110- 155
C (heteroaromatic)
105 - 165
-ON
115 - 125
C=0 (acids, acyl halides, esters, amides)
155 - 185
C=0 (aldehydes, ketones)
185 -225
In 13C NMR spectroscopy the l3C signal due to the carbon in CDC13 appears as a triplet centred at 5 77.0 with peak intensities in the ratio 1:1:1 (due to spin-spin coupling between 13C and 2H). This resonance serves as a convenient reference for the chemical shifts of 13C NMR spectra recorded in this solvent. Table 6.4 gives characteristic 13C chemical shifts for some s/Ahybridised carbon atoms in common functional groups. Table 6.5 gives characteristic 13C chemical shifts for some .syr-hybridised carbon atoms in substituted alkenes and Table 6.6 gives characteristic l3C chemical shifts for some .s/j-hybridised carbon atoms in alkynes.
240
Figure 6.2
220
200
180
160
140
120
100
80
60
Approximate 13 C Chemical Shift Ranges for Carbon Atoms in Organic Compounds
40
20
0
-20
5 (ppm)
Chapter 6 13C NMR Spectroscopy
77
Chapter 6 13C NMR Spectroscopy
CH3 CH2-
c h 3— X
AA
13C Chemical Shifts (5) for sp3 Carbons in Alkyl Derivatives (CH3 )2 CH-— X
— ch3
— ch3
— H
-2.3
7.3
7.3
15.4
15.9
— ch=ch2
18.7
13.4
27.4
22.1
32.3
— Ph
21.4
15.8
29.1
24.0
34.3
— Cl
25.6
18.9
39.9
27.3
53.7
— OH
50.2
18.2
57.8
25.3
64.0
— och3
60.9
14.7
67.7
21.4
72.6
— OCO-CH 3
51.5
14.4
60.4
21.9
67.5
30.7
7.0
35.2
18.2
41.6
20.6
9.2
27.2
19.1
34.1
— nh2
28.3
19.0
36.9
26.5
43.0
— NH-COCH3
26.1
14.6
34.1
22.3
40.5
— CEN
1.7
10.6
10.8
19.9
19.8
— no2
61.2
12.3
70.8
20.8
78.8
CO
X
X
0 1 0 0
X i ro 1
Table 6.4
— CO-OCH 3
Table 6.5
— ch3
13C Chemical Shifts (8 ) for sp2 Carbons in Vinyl Derivatives: CH 2=CH-X X
1 X
II
115.9
136.2
- C ( C H 3 )3
108.9
149.8
— Ph
112.3
135.8
116.3
136.9
129.2
117.3
128.0
137.1
— CO-OCH 3
130.3
129.6
— Cl
117.2
126.1
—o c h 3
84.4
152.7
—0 C0 -CH3
96.6
141.7
— CEN
137.5
108.2
—n o 2
122.4
145.6
— N(CH3)2
91.3
151.3
CM
II
X
— ch3
0
123.3
X
123.3
0
—H
1
X
0
III
1
0
CO
X
0 1 0 0
78
0
c h 2=
X
>
H~
Chapter 6 13C NMR Spectroscopy
Table
13C Chemical Shifts (8 ) for sp Carbons in Alkynes:
6 .6
X-C=C-Y
X —c = H—
—H
H—
—
66.9
79.2
67.0
92.3
80.0
82.8
66.3
67.3
77.1
83.4
81.8
78.1
o c h 2c h 3
22.0
88.2
ch3
72.6
72.6
Ph
79.7
85.8
97.4
87.0
89.4
89.4
74.6
74.6
ch3
c h = ch2
H—
1
—
I
III
H—
0
0
H—
Ph
—coch3 —
c h 3—
—
c h 3—
I
0
73.2
- C ( C H 3)3
H—
H—
73.2
—
H—
—
1CO
—coch3
Ph — CO O CH3
—
—
= C -Y
Ph
cooch
3
Table 6.7 gives characteristic 13C chemical shifts for the aromatic carbons in benzene derivatives. To a first approximation, the shifts induced by substituents are additive. So, for example, an aromatic carbon which has a -N O 2 group in the para position and a -B r group in the ortho position will appear at approximately 137.9 ppm [(128.5 + 6.1(p-N02) + 3.3(o-Br))].
79
Chapter 6 13C NMR Spectroscopy
Table 6.7
Approximate 13C Chemical Shifts (8 ) for Aromatic Carbons in Benzene Derivatives Ph-X in ppm relative to Benzene at 128.5 ppm (a positive sign denotes a downfield shift)
8
X
ipso
ortho
meta
para
— H
0.0
0.0
0.0
0.0
— no2
19.9
-4.9
0.9
6.1
2.0
1.2
-0 . 1
4.3
5.0
- 1. 2
0.1
3.4
8.9
0.1
- 0. 1
4.4
— CEN
-16.0
3.5
0.7
4.3
— Br
-5.4
3.3
2.2
- 1.0
8.9
-2.3
-0.1
-0 . 8
— Cl
5.3
0.4
1.4
-1.9
— ch3
9.2
0.7
-0. 1
-3.0
— o co -C H s
22.4
-7.1
0.4
-3.2
— och3
33.5
-14.4
1.0
-7.7
— nh 2
18.2
-13.4
0.8
- 1 0 . 0
-C O -O C H 3 CNJ
z
X
1
1
0 0
CO
X
O
1
O O
CNJ
X
O
II X
0
Table 6.8 gives characteristic shifts for 13C nuclei in some polynuclear aromatic compounds and heteroaromatic compounds. Table
6 .8
Characteristic 13C Chemical Shifts (8 ) in some Polynuclear Aromatic Compounds and Heteroaromatic Compounds
80
7 2-Dimensional NMR Spectroscopy A two-dimensional NMR spectrum has two frequency axes rather than one. A 2D spectrum is acquired using a pulse sequence which contains a delay period 'tf which can be varied systematically as the experiment is repeated. The experiment is repeated many times (typically 512 or 1024), with a different delay V in the pulse sequence for each experiment. One FID is acquired for each experiment giving an array of'N ' individual FID's each of which has been acquired with a slightly different pulse sequence. Each FID represents the variation of detected signal as a function of time (t2 in the diagram below) and successive FIDs in the array differ as a function of the time variable ti within the preparation period of the pulse sequence. Fourier transformation of the two-dimensional array of data with respect to t2 affords a series of spectra which vary systematically as a function o f f . A second Fourier transformation, this time with respect to ti, gives a two-dimensional spectral array (which is function of two frequency domains F t and F2). Two-dimensional spectra are usually represented in terms of a stacked plot or contour plot. The contour plot is a more convenient representation for making measurements or peak assignment.
Time dependent pulse sequence
time t2
frequency F2
FID
frequency F2
Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
81
Chapter 7 2-Dimensional NMR Spectroscopy
The number of possible two-dimensional experiments is essentially unlimited. Different pulse sequences in the preparation period give rise to different twodimensional spectra which can be tailored to exhibit various properties of the sample. The technical detail behind multi-dimensional NMR experiments and the pulse sequences used to generate 2D spectra, is beyond the scope of this book.
Two-dimensional spectra have the appearance of surfaces, generally with two axes corresponding to chemical shift and the third (vertical) axis corresponding to signal intensity. s h ift ( f T)
It is usually more useful to plot twodimensional spectra viewed directly from above (a contour plot of the surface) in order to make measurements and assignments.
CF®
to
ra
o E Q) . O
Chemical shift (F2 )
The most important two-dimensional NMR experiments for solving structural problems are COSY (C o rrelation SpectroscopY), NOESY (Nuclear Overhauser Enhancement SpectroscopY), HSQC (Heteronuclear Single Quantum Correlation) or HSC (Heteronuclear Shift Correlation), HMBC (Heteronuclear Multiple Bond Correlation) and TOCSY (TOtal Correlation SpectroscopY). Most modem high-field NMR spectrometers have the capability to routinely and automatically acquire COSY, NOESY, HSQC, HMBC and TOCSY spectra.
82
Chapter 7 2-Dimensional NMR Spectroscopy
7.1
COSY (CORRELATION SPECTROSCOPY)
The COSY spectrum shows which pairs of protons in a molecule are coupled to each other. The COSY spectrum is a symmetrical spectrum that has the 'H NMR spectrum of the substance as both of the chemical shift axes (Fi and F2). It is usual to plot a normal (one-dimensional) NMR spectrum along each of the F 1 and F2 axes to give reference spectra for the peaks that appear in the two-dimensional spectrum. A COSY spectrum of 1-iodobutane (CH 3CH 2CH2CH2I) is given below (Figure 7.1). F2 Axis
1H NMR Spectrum 4
J
J ____ k
O (D Q_
CO
(Z
(J) x
<
Figure 7.1
*H COSY Spectrum of 1-Iodobutane (CDCI3 solvent, 298K, 600 MHz)
The COSY spectrum of 1-iodobutane has a set of 4 peaks on the diagonal as well as peaks that are off the diagonal. The COSY spectrum is always symmetrical about the diagonal - the off-diagonal peaks above the diagonal are mirrored on the lower side of the diagonal. The off-diagonal peaks are the important signals, since these occur at positions where there is coupling between a proton on the F 1 axis and a proton on the
83
Chapter 7 2-Dimensional NMR Spectroscopy
F2 axis. The protons which are part of the -CH2I group (Hi) are easy to identify since the halogen substituent characteristically moves these to about 8 3.2 ppm. There are 3 off-diagonal peaks on each side of the diagonal - one indicates the coupling between Hi and H2; the second indicates coupling between H2 and H3 and the third indicates coupling between H3 and H4. If you can identify one of the proton signals, then you can identify the protons that are coupled to it and then work sequentially along a coupled network until all the protons which are coupled together are identified. Hi - * - H 2
I
H3—*-H 4
H I1
H I2
H H I3 U
H
H
H H
C— C— C— C—H O ' h 2- *
- h3
In a single COSY spectrum all of the coupling pathways in a molecule can be identified.
7.2
THE HSQC (HETERONUCLEAR SINGLE QUANTUM CORRELATION) OR HSC (HETERONUCLEAR SHIFT CORRELATION) SPECTRUM
The HSQC spectrum or the HSC spectrum are the heteronuclear analogues of the COSY spectrum and these experiments identify which protons are directly attached to which carbons in the molecule. The HSQC spectrum has the 'H NMR spectrum on one axis (F2) and the BC spectrum (or the spectrum of some other nucleus) on the second axis (Fi). It is usual to plot a normal (one-dimensional) 'H NMR spectrum along the proton dimension (the F2 axis) and a normal (one-dimensional) 13C NMR spectrum along the 13C dimension (the Fi axis) to give reference spectra for the peaks that appear in the two-dimensional spectrum. An HSQC spectrum of 1-iodobutane (CH3CH2CH2CH 2I) is given in Figure 7.2. The HSQC spectrum does not have diagonal peaks. The peaks in an HSQC spectrum occur at positions where a proton in the spectrum on the F2 axis is directly coupled to a carbon in the spectrum on the Fi axis via a 1-bond C-H coupline.
84
Chapter 7 2-Dimensional NMR Spectroscopy
1H NMR Spectrum
F2 Axis
4
J
2
3
k
A
ppm
E
5
o
< D Q_ CO
10
C£ z:
o
15
CO
20
25
30
c/) x <
Figure 7.2
35
*H - 13C HSQC Spectrum of 1 -Iodobutane (CDCI3 solvent, 298K, ’H 600 MHz, 13C 150 MHz)
1-Iodobutane has 4 carbon resonances and the h3- * - c 3
HSQC spectrum shows 4 cross peaks. Having identified all of the resonances in the !H spectrum, then the resonances in the carbon
Ch
H *W I2 V I 3
H
U C— C— c — H
1C 1
spectrum can simply be identified by the
H
positions of the cross peaks corresponding to
7
H4—*-C 4
each proton resonance, In the HSQC or HSC experiment, all of the protons which are coupled to carbons can be identified. It is usually possible to assign all of the resonances in the ]H NMR spectrum i.e. establish which proton in a molecule gives rise to each signal in the spectrum, using spin-spin coupling information or a COSY experiment, then assign the signals in the l3C spectrum by correlation to the known proton resonances. Note that non-protonated carbons (quaternary carbons) do not give rise to signals in the HSQC or HSC spectra.
85
Chapter 7 2-Dimensional NMR Spectroscopy
7.3
HMBC (HETERONUCLEAR MULTIPLE BOND CORRELATION)
The HMBC spectrum correlates chemical shifts of hydrogen nuclei with carbon nuclei which are separated by two or more chemical bonds. The HMBC experiment is frequently used to assign quaternary and carbonyl carbons which don’t have any directly bound protons so they are “invisible” in the HSQC or HSC experiment. The HMBC experiment is designed to filter out correlations resulting from large C-H coupling constants (120-160 Hz) resulting from protons directly bound to carbon and to select for smaller couplings (around 10 Hz) which are typical C-H couplings over 2 or 3 bonds. The HMBC is a very powerful method for making the connection between two parts of a molecule that may be isolated from each other by a carbonyl, an ether, an ester, an amide or by some other functional group. ^
X
H
\H
______ c — x— ,H X = 0, S,NH
1, It is usual to plot a normal (one-dimensional) H NMR spectrum along the proton
dimension (the F2 axis) and a normal (one-dimensional) 13C NMR spectrum along the 13
C dimension (the F 1 axis) to give reference spectra for the peaks that appear in the
two-dimensional spectrum. A HMBC spectrum of 1-iodobutane (CH3CH2CH2CH2I) is given in Figure 7.3. The HMBC spectrum does not have diagonal peaks. The peaks in an HMBC spectrum occur at positions where a proton in the spectrum on the F2 axis is coupled to a carbon in the spectrum on the Fi axis via a 2-bond or 3-bond CH coupling.
86
Chapter 7 2-Dimensional NMR Spectroscopy
1H NMR Spectrum
F2 Axis
A
J
JL
L
£ 3— L o 0
Q-
(/) q: z
o co
-
CO CD CL (D (/) CO _Q 7 0 0^
45
. 40 — 20
No significant UV
M+’
absorption above 220 nm
74 57
li
k
it
1
1 ___ L-
c 3 h60
2
_J— 1— 1— i— 1— 1— 1__1__1__1— 1__1— 1— 1— 1__1— 1__1__1__1__1__1__1__1__
80
40
120
160
200
240
280
m/e 1
1
■
1
1
1—
f
1
1
•
1
1
1
1
1
1
1
.
1
1
1
1
1
■
1
•
1
•
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
c h 2|
CH3f a ■4
solvent
proton decoupled
hi
.............................................................................
200
.
160
I
.
1
.
120
I
80
40
1 -
0
— 5 (ppm)
• ... -
1H NMR Spectrum (200 MHz, CDCI3 solution) exchanges with D20
_A_
12
11
i
i 10
9
8
7
6
5
4
3
2
1
TMS
0 6 (ppm)
113
Problem 3
4000
3000
2000 1600 V (cm'1)
1200
800
0.0
UV spectrum solvent: ethanol 15.4 mg /1 0 mis path length : 1.00 cm
40
120
80
160
240
200
280
250
200
T
' 1
"1..—
I
'
I
'
350
300 X (nm)
m/e I
'
i
■
1
■
1
'
I
'
1
1
1
i
1
1
1
1
1—
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT CH2| CH3| CHf
proton decoupled
solvent III
1 1
1
1
i
1
200
1H NMR
1
1
160
1
1
1
1
120
1
1
1
1
80
1
1
40
1
0
5 (ppm)
Spectrum
(200 MHz, CDCIj solution)
r 1 J
10
114
8
i
L
0 6 (ppm)
Problem 4
4000
3000
2000
1600
1200
800
V (cm'1) Mass Spectrum
57
29 ■ 0 -. Q 03 3. ~ Q 3 CO -. _0Q 3 7O - sS
No significant UV
00 0 0
II +
2
absorption above 220 nm
C 4H 8O2 L ii i » 1 j—i_i_J ____1— 1___ *-- 1-- *-- »-- *-- 1-- 1-- »-- •-- 1-- •-- 1-- *-- •-- •-- 1-- •-- 1—- i • * *
40 1
80 1
'
120
' ... 1
1
160 m /e
1
' 1
200 '
240
1
'" " " T
280 1
1
'
,
1
.
|
1
|
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
1
..................
DEPT
CH2(| CH3f C H f
proton decoupled
solvent I
I
■
I
■
I
200
10
9
■
I
.
I
160
8
7
■
1
■
1 ..... X
120
6
5
1
■
1 — I------- \-------1------ 1------- 1-------1-------1----------------
80
4
40
3
2
0
1
S(PPm)
0 6 (ppm)
115
Problem 5
1600 2000 V (cm'1)
3000
4000 100 :
Mass Spectrum
27
80 60
800
1200
107/109
- ^ • CO - 0 . Q_ “ 0
No significant UV
1 0 . _Q
40 r
absorption above 220 nm
o
M+ ’
- osO N
186/188/190
20
,i, C 2 H 4 B r2 1 U, At. .1 i, ■ » *__I__.__I_____I___—I__1—«— 1—I— 1—I— 1—I—*—i—•— 1— 1— 1—>— 1—■— 1—•— 1—'—
120
80
40
160
240
200
280
m/e ------------- I—
'—
i—
i
>-
1
i
>
i
1
i
i
■
i
■
i
■
i
1
. 1
'
1
'
.
1
.
"1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
solvent |
proton coupled
proton decoupled
I
.
I
^
*
1
1
1
200 1H NMR
1
1
160
-1
J ___ j
1
1
120
L
80
40
I
0
5 (ppm)
Spectrum
(200 MHz, CDCI3 solution)
TMS
...............
j
I
10
116
1
I
9
I
»
8
I
1
7
1
- t ------ 1-------- 1-------- 1 ----- -—1-------- 1-------- •-------- 1 -------- •-------- 1-------- 1-------- 1 -------- 1-------- 1 -----
6
5
4
3
2
1
0 8 (ppm)
Problem 6
2000
3000
4000
1200
1600
800
V (cm'1)
100
Mass Spectrum
43
80 -■ CO -. Q CD. D 60 -“ cC n - C O . _Q 40 7 o
UV Spectrum ^max 289 nm (log^gS 1.4) solvent: methanol
C CO O II + 2
20 r i. i C 4 H6 ° 2 __1___1___L » i_u—1_1—1_1_1_1_1_1_1__ _l-- .-- 1-- ,-- 1-- .-- 1-- .-- 1-- .—J- —■—i 280 200 240 120 160 40 80
m /e 1
’
1
1
1
1
1
1
’
1
1
1
|
1
1
1
.. 1
r
1. i
1
13C NMR Spectrum (100 MHz, CDCIj solution)
D EP T
c h 2| c h 3| c h |
proton decoupled
solvent
* I
.... i
i
i
i
i
i
200
i
i
i
i
i
160
i
i
i
i
i
120
.................. i
i
.j
_i-------1-------1-------1-------1-------1-------1----------------
80
40
0
5 (ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
TMS
— I------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1___
10
9
8
7
6
5
1
-
4
•
1
3
1
«
2
1
.
1
1
1
1
0 8 (ppm)
117
Problem 7
4000
3000
2000 1600 V (cm'1)
800
1200
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e --------------i—
i—
\—
i—
r
1
' ...i
1
i
i
r '"
1
i
1
i
1......... i
i
■
i
■
.
I
,
i
1
I
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
c h 2| CH3f
cuf
proton decoupled
I
,
I
I I
.
I
.
I
200
.
I
.
160
solvent 1
I
.
I
.
120
I
.
I
.
I
80
40
0
S(ppm)
I
i
1H NMR Spectrum (400 MHz, CDCI3 solution)
I
10
118
i
I
9
i
I
8
'
I
7
'
I
6
i
5
I
i
4
I
i
3
!
i
2
I
i
1
L
0 6 (ppm)
Problem 8
4000
3000
1600
2000
800
1200
V (crrf1 )
No significant UV absorption above 220 nm
80
40 1 1 , . 13C NMR Spectrum
120 1
,
160 m /e
1
1
1
240
200
,
j
1
280
1
1
1
1
1
1
1
1
■
1
1
( 5 0 .0 M H z , C D C I 3 so lu tio n )
DEPT
c h
2|
C H 3f
CHf
p ro to n d e c o u p le d
I
.
s o lv e n t
I . 200
i---- 1
i i i
i
160
i i----1 120
i
III
I
i . 80
i--- 1---- 1---- 1---- 1---- 1---- 1 _i-------------------40 0 5 (ppm)
1H NMR Spectrum (200
M H z,
CDCIg s o lu tio n )
TM S
i
i
1
1
1
1
1
------ ------------ ------------ ------------1------------ ------------1----------- ----------- 1—
10
9
8
7
6
i
«
5
1
•
4
i
1
3
1
i
2
i
1
1
1
i
1
0 5 (ppm)
119
Problem 9
100 :
42
Mass Spectrum
80 “• CO - CD . Q. CD _ CO . _Q
No significant UV
60 -
40 7
M+55
O
absorption above 220 nm
70
- -5
20 C 5 H 10
: I, 1 ___ ._1_1_L _J-- 1.
40
.
.
1
.
.
1
1
120
80
i-- .-- 1-- .-- 1-- .---- 1-- .---1-- .-- 1-- .-- 1-- .-- 1-- .--
.
200
160
240
280
m/e 1.......' MMR (1 2 5 .0
MHz,
1 ..T ■"" I
1
1
1
1
' .....1
.9 n p r .tr n m
1
1
' ' .... ”T.......r.... 1
1
1
1
1
1
1
'
I l 40
I
1
1 0
l 8 (ppm)
_
C D C I 3 solutio n)
DEPT
CH2{ CH3f C H f
pro to n d e c o u p le d
s o lven t
1
l
I
l
I l 200
1
l
1 l 160
I
1
I l 120
I
1
I i 80
I
l
1H NMR Spectrum ( 4 0 0 M H z , C D C I 3 solutio n)
_J
10
120
L_
-I
I
8
i
L
7
_i
L
1
0 8 (ppm)
Problem 10
4000
1600 V (cm‘1)
3000
800
1200
2000
100 :
Mass Spectrum
59
80 " ro . C0
l
No significant UV
60 * 00 -
40
. _CQO 7o - £5
absorption above 220 nm M +’=118 (
8
1
.
7
1
.
6
1
5
1
.
4
1
.
3
1
.
2
1
i
1
1
1
1
0 5 (ppm)
Problem 54
4000
2000
3000
1200
1600
800
V (cm'1) 100
152
-
Mass Spectrum
“ •
CD in
:
40 7 -
20
76
-0 0
00
CO CD
CM CD
60
■ -
X CM X—
0
80
UV Spectrum
M
A,max 264 nm (log10e 4.5)
+ '
solvent: methanol
310/312/314
•
i | i
-t 1 1
1 .II, Jlil .il 1, l ------ .-- 1-- .---1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- L--. 1 . 1 . 1 . 40 80 120 160 200 240 280
m/e I
I
'
I
'
' 1
1
«
I
|
.
j
,
1 ■ ,
|
.
j
1
|
,
1
.
1
.
|
.
1
.
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2)|f CH3f
ch
|
solvent
proton decoupled
i
•
I
I
I
1
200
-1
1 .. 1
160
1
1
1
.
.
120
1
1
.
1
80
1H NMR Spectrum
40
0
8
(ppm)
expansion
(400 MHz, CDCI3 solution)
residual solvent
7.8
7.6
7.4
7.2
ppm
TMS
-J----- 1----- 1___ I___ I___ 1
- J _________ I_________ I_________ I_________ L_
10
9
8
7
6
5
4
1 3
_J_________ I_________ I_________ I_________ L_
2
1
0 8 (ppm)
165
Problem 55
4000
3000
2000 1600 V (cm'1 )
100 :
1200
800
Mass Spectrum
125/127
8 0 -• CO -
CD
6 0 -■ 5
-. CL CP CP 60 “ ■ C/3 -. -O CO 40 7 o - -s
20 C 12H 6 1 il II il Jl l III J ------ 1---1—1—1—1—1—1_1_1_1_l_l_1_1_■ —i—i—i—i_i_i—i_i_i_i_i_i_i_■ i ■
120
80
40
200
160
240
280
m/e i
i
1
I
•
1
I
1
I -1
1
|
|
.
|
1 .... 1
1
1
1
1
1
1
1
1
*
1
1____I____1_________
13C NMR Spectrum (75.0 MHz, CDCI3 solution)
D EPT
CH2J CH3f C H f
1—v- solvent
proton decoupled
1
l
*
.
I
1
.
200
•
l
160
*
l
1
.
.
120
I
1
•
«
80
40
0
5 (ppm)
1H NMR Spectrum (300 MHz, CDCI3 solution)
f.......
solvent residual
TMS
r 1 .
i
1
10
.
1
9
1
.
8
1
1------- 1------- 1------- 1------- 1—
7
6
5
1
1
4
1
1
3
1
1
2
1
1
1
......1_____..........1 .
0 5 (ppm)
171
A-(nm)
m/ e — .. . r
'
1
i
i
1
i
1
i
^
1
1
1
1
1
'
1
1 .. 1
1
1
1
1
1
1
1
1
1
1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
ch 2| ch 3| c h |
solvent
proton decoupled
1 ----- 1------1----- »----- L----- 1------1----- 1----- 1----- 1------1------1------1----- 1_— 1-----j
200
160
120
1
1
1
80
40
0
8
(ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
___4 TMS L _ I
10
172
1
1
9
1
1
8
—1------- 1------- 1------- 1------- 1 — 1----- 1
7
6
5
4
1
1
3
1
1
2
_
1
1
1
1
1
1
0 8 (ppm)
Problem 62
4000
1200
2000
1600 V (cm'1 )
3000
100-
800
Mass Spectrum
105
UV Spectrum
80 Xmax 261 nm (log10s 2.7) 60 ^max 269 nm (log^s 2.5 )
40
M+,= 120 solvent: methanol
20 C9H12 40
80
120
160
200
240
280
m/e
1---- 1----1----1----1
■ ---- 1----1---- 1----1---- 1----1---- 1----'---- 1----•---- 1---- •---- 1----1---- 1----■ ---- 1----r
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EPT
CH2| CH3f
ch
|
i
solvent
proton decoupled
I
i
10
■ i . 200
9
i
8
■ i . 160
7
i
6
....................
.
i ■ i 120
5
..........1 ................
,
4
i 80
.
3
i
_
■ i 40
2
....... _
■ i
1
1
.
i 0
5
(ppm)
0 8 (ppm)
173
Problem 63
100
105
Mass Spectrum
UV Spectrum
80 A-max 270 nm (log10£ 2 .6 )
60 - 0CO
M + ' = 120
CO
solvent: methanol
40
20 C' 9q n H,12 _J
I
I
I
I
I
I
I
I
40
I
I
i
I
I
I
I
120
80
L_
-I
200
160
■
I
i
I
i
240
I
i
L_
280
m/e 13C NMR Spectrum MHz, CDCIj solution)
expansion
(100.0
11I' ‘1’I11' 1I11 135 130
DEPT
ch 2|
aJII. ' i 1i ' i i ■i ■
PPm
24
CH3f CHf expansion
135 130
JU i " i ■-11i 1i 1
solvent
ppm
24
20
J
200
ppm
expansion
11I■’1’T'... I
proton decoupled
20
160
120
ppm
I
I
80
I
I
1
40
I
0
L.
5 (ppm)
1H NMR Spectrum (400
I 10
174
MHz, CDCI3 solution)
■
I 9
i
I 8
■
I 7
i
I 6
I 5
I
I 4
I
I 3
I
i 2
I
i 1
L 0 5 (ppm)
m /e
I
1 l
1
l
'
l
1
X (n m )
l
1
l
1
I
1
i
i
1
i
1
i
1
1
r
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EP T
CH2| CH3f C H f
solvent proton decoupled
_JL_
.............................. I__ I__ I __ I__ I__ I__ I__ I__ 200 160 120
I ' l l __ I__ I__ I__ 1__ I__ L. 80 40 0 8 (ppm)
1H N M R S p e c tru m (400 MHz, CDCI3 solution)
r
f TMS L ----- 1--------- 1--------- 1--------- 1_____ t_____ L
10
9
8
1
7
1
i 1
1
6
.
5
1
.
4
1
.
3
1
1
2
1
,
1
1
,
1
0 8 (ppm)
175
Problem 65
IR Spectrum
802
(liquid film)
_L
_L
4000
3000
2000
1200
1600
800
V (cm1)
120
80
40
200
160
240
280
m/e T
' '
1.. “ I
1“
.... 1
'
' ~ 1— r_- i
1
r....•
1
1
•
1
•
1
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
CHa| CH3| O l f
solvent proton decoupled
- ___ I—.. 1
I
1 ......*
l
i
l
.
l
________________J L ______ ___ ________
1
t
160
200
1
.
80
120
1
........................................................
40
0
8
(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
r 1
"
TMS
-> u
J 10
176
i
I 9
i
I 8
i
I 7
i
I 6
i 5
I
■ 4
I
i 3
1
—. V
...
I
i 2
I
i 1
L 0 6 (ppm)
Problem 66
2000
1600 V (crn1 )
3000
4000
100
119
1200
800
Mass Spectrum
UV Spectrum
80 -
. CD Q_
^max 274 nm (log10s 2.3)
M+* = 134
60 - 0
^max 277 nm (log^s 2.3)
40
solvent: cyclohexane
20 C 10H 14
-J ■ --1--1--1--.--1-- I--1--1_ I_._I___I_............... 40
80
120
160
200
240
280
m/e *C NMR Spectrum (100.0 MHz, CDCI3 solution) expansion
DEPT
130
140
ppm 20
CH2)|f CH3f C H f
expansion
l
■ I 200
ppm
expansion
_ k jL proton decoupled
15
solvent
140
130
I
I 160
ppm
20
120
80
15
ppm
40
0
5 (ppm)
5 (ppm)
177
Problem 67
UV Spectrum max
278 nm
2.4)
X max 274 nm (log10s 2.4) A,max 270 nm (log10s 2.5) solvent: methanol
120
80
40
200
160
240
280
m/ e ----------- r ..1 i ' 13C NMR Spectrum
1
i
i'
■
i
•
1
i
1
i
,
r
|
1 "7
1
, ___1
'
1
1
1
1
,
1
1
(50.0 MHz, CDCI3 solution)
proton coupled
I
sol'vent
j
proton decoupled
I
.
I
1
.
.
I
200
.
I
160
i - .... 1..... 1
l
i
l
120
,
80
40
0
S(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
TMS .
I
10
178
1
I
9
1
I
8
1
I
7
J--------- 1--------- 1----
1
6
5
l
J
J--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1-----
4
3
2
1
0 8 (ppm)
Problem 68
UV Spectrum
X max 232 nm (log10s 3.4) ^max 248 nm (log^E) 3.5)
^max 265 nm (log^s 3.4) solvent: isooctane
40
120
80
160
200
240
280
m/e 1
1
1
1
T
1
1
'
1
'
1
'
1
'
7 ... t
1
1------- I
13C NMR Spectrum
1
1
'
1
1
expans ion
( 100.0 MHz, C D C I3 solution)
J
1.
1 1 1 1.... 1 D EP T
15
C H 2j CH3f c h |
13
ppm
expans ion
solvent i
proton decoupled
15
___________
1
.
1
200
.
1
.
1
160
1
_________
1
...
.L
T--- T---T,,,1---- 1
i
«
l
120
i
l
1
.
1
80
___ .... 1
1
13 1
40
ppm 1
1___.___ 1___ 1_______
0
s (ppm)
8 (ppm)
179
Problem 69
No significant UV absorption above 210 nm
40
120
80
200
160
240
280
m/e "I— 1----1— '— T 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
proton coupled
X solvent proton decoupled
i
10
180
120
160
200
8
6
80
40
0
1
5 (ppm)
0 8 (ppm)
Problem 70
4000
3000
2000
1200
1600
800
V (cm'1)
No significant UV absorption above 220 nm
40
120
80
160
200
240
280
m/ e 1
1
I
1
1
1
'
r"
1....... 1
' 1
1
1
1
'
'
"
1
'
1
..'
1
'
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
w
D EP T
nr, -r™,
C H C H 3f CHf
solvent proton decoupled
.
- - ....... -....-...... .. « 1 1 1
1
I ____ _____............ _......... ........... _ 1----- •- 1----- 1 1 1 1 1
200
160
...... I L _
1
120
1
1
1
1----- 1----- 1----- 1------ 1--
80
40
1 - 1
0
6
(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
exchanges with D20
TMS
J 10
i
I 9
i
I 8
,
L 7
-I_____ I_____ I_____ I_____ l_
6
5
4
J
i 3
I
i 2
I
i 1
L 0 6 (ppm)
181
Problem 71
4000
3000
2000
1600
1200
800
V (cm'1)
120
240
160
280
m/e i ..
'
i
1
r " ... 1
r "
X (nm) t
1
1
1
1
'
1
1
1
1
1
1
1
• .....r
.
1
.
1
,
1
,
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
ch
2| C H 3f c h |
proton decoupled
solvent
1 -1
-L -
*
J___ j
200
1
,
1
160
,
1
.
1
120
.
1
.
1
80
40
1
0
8 (ppm)
8 (ppm)
182
Problem 72
2000
3000
4000
1200
1600
800
V (cm
100-
Mass Spectrum
No significant UV absorption above 220 nm
c 5 h7 n o2 120
80
40
200
160
240
280
m/ e ....'.......1........'..... ..1.... —r ------ r................1....... ' ....... 1
1
I
•
I
»
I
1
'
1
•
'
r
_
i
1
i
1 ............
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f C H f
proton decoupled ■
i
i
i
I
i
i
200
i
i
i
i
i
160
I
solvent
1
i
i
i
i
i
120
i
. i
i____i____ i____ i
80
______ 8 (ppm)
__i____ i____ i
40
0
1H NMR Spectrum (200 MHz, CDCI3 solution)
r -
,---------j TMS
|
................................ .......................... — 1—
10
1—
1—
9
1—
1—
8
1—
1—
7
.
1____ 1____ 1...... 1
6
5
1
4
1\ 1
1
j
1
..... 1........ 1...... 1
3
2
1 .1
1
j
. 1
0 8 (ppm)
183
Problem 73
4000
2000
1200
1600 V (cm 1 )
3000
800
No strong UV absorption above 220 nm
120
80
40
160
200
240
280
m/e ■— | - 1
|
.
|
1
|
.
|
!
b
1
'
1
,
.
1
1
y— .
|
,
1
1
1
,
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
ch 2| ch 3| c h |
proton decoupled
solvent ill
I
--------- 1----- »----- 1----- 1----- 1----- *----- 1----- «------1----- »----- 1----- 1------1----- 1----- 1----- L
200
10
184
9
160
8
7
120
6
5
i
80
4
40
3
2
1
1
1
0
6
(ppm)
0 5 (ppm)
Problem 74
UV Spectrum
X max 283 nm (log10s 1.4) solvent: hexane
40
80
120
160
200
240
280
m/ e i — 1— r 13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
ch 2|
CH3f
ch|
proton decoupled solvent
* 1 I
J
1
I 200
1
I
1
L 160
_L 80
120
J
L.
0
40
8 (ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
4.0
J 10
1-----
3.0
exchanges with D20
2.0
ppm
1----- 1___ I___ 1____I___ i____> 8
6
■
l
■
l
■
l
1
I 1
0 8 (ppm)
185
Problem 75
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e 13C NMR Spectrum (100 MHz, CDCI3 solution)
D EP T
CH2^ CH3f CHf
solvent
proton decoupled
> «__ I__ 1__ I__ 1__ I__ 1__ I---- 1----1---- 1---- 1__ 1__ 1__ 1__ I__ i__ I__ 1__ I__ 1 200 160 120 80 40
I__ 1______ 0 § (ppm)
8 (ppm)
186
Problem 76
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e ,3C NMR Spectrum (100 MHz, CDCI3 solution)
DEPT
ch 2|
CH3f
ch|
solvent
proton decoupled _J
I
I
200
1
I
I
I
160
»
I
I
»________1
120
I________ t
i
80
l
l
!
1
________ I________ |________ I________ |________ ________ L_
40
0
5 (ppm)
5 (ppm)
187
Problem 77
2000
3000
4000
1200
1600
800
V (cm1)
100;
Mass Spectrum
80 No significant UV
60
absorption above 220 nm
40
20
M+* = 88 (< 1%) C5 H1 2 °
120
80
40
200
160
240
280
m/ e '
i
i
■r” ...r
1
i
1
1
i
1
i
i
1
1
i
i
1
i
1
1
1
i
1
1
1
r
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
CH2i|f CH3f C H f
proton decoupled
solvent^
____ 111 1 ____ 1 1 1
— 1------1------1------1------1------1------1----- 1----- «------1------1----- 1----- «------1----- •----- 1— 1
200
10
188
9
160
8
7
120
6
80
5
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
Problem 78
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e t—
1— I— •— r
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
CH2(i CH3f CHf
solvent proton decoupled
J
200
1
j
L
160
120
I
»
I
80
.
I
1
I
40
1
I
L
0
5 (ppm)
5 (ppm)
189
Problem 79
4000
3000
2000 1600 V (crrf1 )
1200
100;
800
Mass Spectrum
80 No significant UV
60 40
absorption above 220 nm
107/109
M+'
20
152/154
180/182
C5H9 ° 2 Br »
.
I
.
i
i___ ____ I___ ____ I___ ____ I___ i___ L .
.
40
80
120
160
200
240
280
m/ e i— ■ — I— 1— r 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
ch 2| ch 3|
CHf
solvent proton decoupled
J
.
I . 200
I
1
L 160
J
J
L.
120
.
I I 80
\__
I 40
-I
I 0
I—
8 (ppm)
6 (ppm)
190
Problem 80
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
CH2J CH3f C H f
proton decoupled solvent
J
I
I
i
I
I
I
I
I
I
I
.
I
,
120
160
200
I
.
I
I
80
I
L.
J
40
L-
0
5 (ppm)
i
1
1H NMR Spectrum (400 MHz, CDCI3 solution) expansion
5.0
J
10
expansion
4.5 ppm
3.5
3.0
2.5
ppm
1------- 1------- 1------- 1------- 1------- 1------- 1____ I____ 1____ >
8
1
1
■
i
■
i
»
1
0 8 (ppm)
191
Problem 81
4000
1200
2000
1600 V (cm'1 )
3000
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/ e 1
1 1 ' "I..... 1 13C NMR Spectrum
,
,
.
,
-|
.
|
.
,
.
|
.
•
1
>
1
1
1
1
1
•
1
.
1
'
1
1
(50.0 MHz, CDCI3 solution)
D EP T
c h 2|
c h :it
CH f
solvent proton decoupled
-
-1
-i— 1
L ............................................. • -...1 1 I -l 1 —1___ L..
200
10
192
9
160
8
7
1 1
1
120
6
5
1
80
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
Problem 82
4000
100
3000
:
60
. •
800
Mass Spectrum
69
41
80
1200
2000 1600 V (cm'1 )
^ CO CD
97
Q.
CD CO
No significant UV
115
r .Q 40
7 _
absorption above 220 nm
o M
+ ’ =
160
(<
1%)
20 L i
i
iiii
_________ i____1___ j ____i
It!
uU
_ i ____ i_ _ _ l
_
i
.
120
80
40
, ( 7 — 1------*-- 1------■------•------■----- «
*
c 7 h 12o 4 -- •-- L . 1 * 1 *
-- 1
<
*
-•------1
200
160
240
280
m/e 1
1
■
1
■
;
1
1
1
1
1
•
1
1
1
1
1
1
■1
r .•
■■■■■'
1 ........>
13C NMR Spectrum (125.0 MHz, CD3OD solution)
DEPT
CH2ijf CH3f C H f solvrent
proton decoupled
^| ................
I
1
1
I
•
1
200 1H
1
1
160
.
1
.
1
.
120
NMR Spectri
1
1
.
1
.
80
.
I
.
40
TMS 1 I
0
1
8
(ppm)
2 protons exchanged for deuterium in this solvent
(600 MHz, CD3OD s•olution)
expansion
1
r ‘ 2.2
1 1.8 ppm
solvent residual
solvent residual
i ----- 1--------- 1---------1--------- 1--------- 1---------1--------- 1---------1--------- 1---------1---------1.......... 1
10
9
8
7
6
5
4
1
■
3
r
____| ___
l
l
1
I
2
TMS
I
■
1
1
1
I
0 8 (ppm)
193
Problem 83 A Ir'
IR Spectrum (liquid film)
i
.
i
^
i
.
i
.
»
»
.
1640 / II « ■ I_______.___,___I_______I_______L
1200
2000
1600 V (cm'1 )
3000
4000
.
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e i
■
1
i
1 ......r "
i
1
i
i
•
1
i
1
i
1
i......1 ........ i
|
1
i
1-
13C NMR Spectrum (125.0 MHz, CDCI3 solution)
D EPT
c h 2|
c h 3|
ch
|
solvent proton decoupled
|
TMS
^
1 ------ 1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1------ 1
200 1H
160
120
1
1
1
1
80
1
1
1
1
40
1
0
5 (ppm)
NMR Spectrum
(600 MHz, CDCI3 solution)
TMS
1 1
10
194
*
1
9
1
1
8
1
1
7
1
1
J___1
6
5
.
i
4
1
.
3
.
2
1
.
1
1
1
1
0 5 (ppm)
Problem 84
1200
2000
1600 V (cm*1 )
3000
4000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e
200 1H
160
120
80
40
0
8
(ppm)
NMR Spectrum
(400 MHz, CDCI3 solution)
expansion
expansion
I 3.8
3.6
ppm
I
- 1------ 1— 1 ..T. 1.6 ppm 1.8
■ —
TMS
1 ----- 1--------- 1--------- 1--------- 1--------- 1--------- 1---------1---------1---------1---------1---------1---------1—
10
9
8
7
6
5
4
j L
1
1
3
1
I
2
L 1
I
1
1
1
1
0 5 (ppm)
195
Problem 85
No significant UV absorption above 220 nm
40
120
80
200
160
240
280
m /e .............1 ' 1 1 13C. NMR .^nAr.tri im
1
1 T " .... '
1
QEPT
(100.0 MHz, dioxan-D8 solution)
1
1 ”r
1
'
1
1
1
i
.
I
1....... 1
l
■
I
.
l
CH2| CH3f C H f
proton decoupled
I
.
I
.
I
.
I
.
I
.
160
200
i
.
.
.
I
40
80
120
.
.
I 0
. 8
(ppm)
1H NMR Spectrum (400 MHz, dioxan-D8 solution)
TMS
L I
10
196
i
1
•
I
8
i
1
i
—I
-i------- 1
—i
1
i
i_
1-------1—
1------- 1------- 1------- 1------- 1----
1
0 5 (ppm)
Problem 86
1200
2000
1600 V (cm"1 )
3000
4000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e 1...... '
1...
1 .. 1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
,
D EP T
.
I ch 2(i
.
1
I
1
1
'
1
'
I
-
1
'
1
'
I
.
I
«
1
CH3f CHf
solvent proton decoupled
I
■
I
u .
i
200
.............................................
160
i
120
•
80
40
I
i
0
8
(ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
TMS
I ---- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1— -L -
10
9
8
7
6
5
1
4
1
1
3
1
1
2
1
I
1
1
1
1
0 8 (ppm)
197
Problem 87
4000
3000
1200
2000 1600 V (cm"1 )
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e 1
------- r 1 i 1 .................................. ....... 13C NMR Spectrum
|
!
|
*
|
1 " 1..... 1
1
1
1........ ..
1,
1........._
(50.0 MHz, CDCI3 solution)
proton coupled
j. . . . .
j
ta.___________
_______ i
solvent proton decoupled
i
I .... 1..... 1
............. L_ .
1
I
200
I
1
ill
1
160
1
.
1
.4.. J..... 1
120
1
1 .. J ....... 1
80
1 ,1
40
0
5 (ppm)
1H N M R S p e c tru m
(200 MHz, CDCI3 solution)
r
expansion 1
l
with resolution enhancement
1 4.2
1............r.. 5.4 ppm
....I
5.8
10
198
...........i.. -
9
1
1
8
1
,
7
1
,
6
1.j V.
V
jL
I
expansion
with resolution enhancement
,
5
1
T
4.0 ppm TMS 1
.
I
4
...........
1
3
1
.
.....L........._L
2
1
1
.1_____L
0 8 (ppm)
Problem 88
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
n— t1— 1r
1----'---- 1----'----1---- T-------- '----1----1----1----r
r-
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
C H C H 3f CHf
solvent
proton decoupled
I
10
.
I
.
1
8
I____ 1____ I____ 1____ I____ 1____ I____ 1____ I____ 1____ I____ 1-------L
J
1
L_
0
5 (ppm)
199
Problem 89
No significant UV absorption above 220 nm
m/e
— r
1— 1— I— 1— I— r 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
ch 24
CH3f
ch|
solvent proton decoupled
J
200
!
I I
I
I
160
I
L
-1
I
120
L.
J
I
1 I 80
I
I
I
40
I
L
0
S (ppm)
8 (ppm)
200
Problem 90
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
120
160
200
0
40
80
5 (ppm)
1H NMR Spectrum (300 MHz, CDCI3 solution)
expansions
Ji 3.8
3.7 ppm
2.1 2.0 ppm
r
1.3 1.2 ppm
TMS
J
10
1
I
i
I
8
i
I
■
I
6
A_________L -i__ i____ i_____i__i_____i__i_____i__ i i 4
3
2
1
0 5 (ppm)
201
Problem 91
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e 1
1
1
I
'
I
1
I
'
1
1
I
1
I
'
1
'
I
'
.
I
.
I
'
1
'
I
.
I
.
1
1
'
I
1
13C NMR Spectrum (125 MHz, CDCI3 solution)
D EPT
c h 2|
C H 3f c h |
solvent -fo
proton decoupled
I
,
I
200
.
1
.
1
160
.
I
.
I
120
.
1
.
1
80
40
0
5 (ppm)
8 (ppm)
202
Problem 92
100
55
Mass Spectrum
82
80 60
• -
No significant UV
CD CO 03
absorption above 220 nm
40
20
M+' = 83
C5H9N
— 1-1--.--1_1_I_1_I_1_I___I..... ............................... -I 1--1-- 1--.--1— 40
80
120
160
200
240
280
m/ e I ‘ I ' I 13C NMR Spectrum (100.0 MHz, CDCI3 solution)
'
1
I
I
'
I
1
I
'
I
1
I
1
I
1................
solvent ^
______ ______________ _______________________ I .................................................................... .......
200
1
I
CH2{ CH3f C H t
proton decoupled
.
I
|
D EPT
1
1
160
120
.
80
_ j— ,— 1— 1— 1— ,------------
40
0
5 (ppm)
6 (ppm)
203
Problem 93
100 ;
Mass Spectrum
43
80 No significant UV
60 40
71
absorption above 220 nm
M+’ = 86 (< 1%)
20 c 5Hi 0o _i
.
i
.
i
i
40
i— •— i— .— i— i— i— *— i—
120
80
160
200
240
280
m/ e -------- r
1
i
1... i
1
i
^
i
•—
i
•
i
'
1
t
r
1
nr
«
1
'
1
i
1
i
1
1
1
i
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
CH2i|f CH3f C H f
solvent
proton decoupled
111
----------- 1-------1-------1------j-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1_
200
10
204
9
160
8
7
120
6
5
i . l
i
80
4
40
3
2
0
1
8
(ppm)
0 5 (ppm)
Problem 94
1200
2000
1600 V (cm'1 )
3000
4000
800
UV spectrum 49.5 mg /1 0 mis path length : 1.00 < solvent: hexane
i . . . . . . . .. i .... 120
80
40
200
160
240
200
280
250
m/e 1
1
I
■
I
■
1
■
300
.
350
X (nm) 1
■
1
1
1
1
1
1
1
1
1
1
1
1
1
1 ....r-
1
1
1
1
1 —.....-
13C NMR Speictrum (100.0 MHz, CDC l3 solution)
D EPT
c H2(,
C H 3f
ch
|
proton decolupled solvent 1 1
I
1
1
1
1
200
1
1
1
1
160
1
1
1
1
1
120
1
1
80
40
1H NMR Spectrum
1
0
1
8 (ppm)
expansion scale expansions
(400 MHz, CDCI3 solution)
0
20
Hz
2.2
2.3
TMS
-J 10
1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1___ 1___ I___ 1___ I___ 1___ I___ 1 9
8
7
6
5
4
3
2
1
1
1
1
0 8 (ppm)
205
Problem 95
1200
2000
1600 V (cm'1 )
3000
4000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e ..I
1
I .... J
1
i
i
•
1
i
i
*
i
■
i
1
I
'
.
I
.
I
1
I,
.
I
,
I
,
,
.
13C NMR Spectrum (100.0 MHz, D2 O solution)
D EP T
ch 2|
CH3f CHf
proton decoupled
I
.
I
.
I
200
.
I
.
I
.
160
I
.
I
,
120
I
80
40
I
0
.
5 (ppm)
1H NMR Spectrum expansions
(400 MHz, D2 O solution) Note: there are 4 protons which exchange with the D2 O solvent
t-------1------ 1-------r 3.4 ppm
1.3
1.1
ppm
H2O and HOD in solvent
■1 J
10
206
9
8
7
1
I
L
6
5
■ 4
l
1 3
I
J
2
1
1
L
0 6 (ppm)
A-(nm)
m/ e I
1
I
.
I
.
I
.
I
1
.
'
1
'
1
1
1
'
1
1
1
'
1
'
R e s o lv e s into
13C NMR Spectrum
tw o sig n a ls a t h ig h e r field
(5 0 .0 M H z , C D C I 3 so lu tio n )
ch2\ C H 3f
D EP T
c h | i
R e s o lv e s into
i
r
tw o s ig n als a t h ig h e r field e x p a n s io n s _ I p ro to n d e c o u p le d .................................................................... ! I
1
1
1
1
200
10
9
1
1
1
..... 1
160
8
I
s o lv e n t
7
_-
.
........
»
1
1
. _
Ui 1
1
120
6
5
3 5 .0
1
l
______ L _
I
'
'
3 4 .0
ppm
_____ 1
1
1---------1--------- 1---------1--------- 1---------1--------- 1--------------------
80
4
i I
1
40
3
2
0
1
5
(ppm)
0 5 (ppm)
207
m/e T
X(nm)
'----1----'----1----'----I----1---- 1----1----1----1----1----'----1----'----1----'----1----1----1---- 1----1---- r
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
c h 2|
c h 3|
CHf
r- .
J
i
I
200
.
H
solvent
proton decoupled
I
i
I
160
i
I
r
i
. — .........
I
120
i
I
r .
i
I
80
____
___________ J
i
I
i
I
40
i
I
i
I
0
i________I 8
(ppm)
8 (ppm)
208
Problem 98
4000
2000
1200
1600 V (cm'1 )
3000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
200
160
120
80
40
0
8 (ppm)
8 (ppm)
209
m /e |
1
|
—T —
1
|
|
,
X (nm) ,
|
.
|
.
|
i
, ....r "....r™... 1......i...
1......r......■.............
expansion
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
i .........1-------- 1— 140 130 ppm
i D EPT
CH2|
c h 3|
ch
|
expansion
I
140 proton decoupled
----- »----- 1----- •----- 1----- *----- 1----- 1----- 1----- •------1------1------1----- 1----- 1----- •
200
10
210
9
160
8
7
120
6
5
130 ppm ryv- solven't
1
i
I
i
I
80
4
i
I
40
3
2
i
l
0
1
i
5 (ppm)
0 5 (ppm)
Problem 100
UV Spectrum ? lmax 262 nm (log10s 2 .5) solvent: ethanol
40
120
80
160
200
240
280
m/ e |
r
|
,
|
,
|
,
|
,
l
'
I
1
I
I
'
I
.
I
.
I
,
l
.
I
.
13C NMR Spectrum (100.0 MHz, CDCIj solution)
expansion
i
-i------------------- 1— 130 125 ppm DEPT
ch 2|
CH3f CHf expansion
i
I 130
l
.
I
200
125 ppm r->- solvent
proton decoupled .
I
.
I
160
,
l
.
I
120
.
I
,
I
80
,
I
i .
I
40
0
5 (ppm )
5 (ppm)
211
Problem 101
100 r
Mass Spectrum
105
80 -■ CO - 0 0 60 "■ C - CO O -Q 40 7 o -
UV Spectrum ^max * 260 nm (log^s « 2.5)
00
II -i-
20 II 1 II __ 1_1_1___1_I_ 1 - 1 40 80
C 10H 14
lI
1 —1_1_._1_._1___1_1_I—* -«—* J * • *-- •-- *-- 1-- * 120 160 200 240 280
m/e 1
T..... 1 "1
1
1 "H..... '
1
1
'
1.. '
1
1
1
1 ... i
1
i
i
.■
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f C H f
1
1
solvent proton decoupled
t i
I ....
I
1
.
1
,
200
10
212
9
.
1
. .... 1
160
8
7
. 1 _ ,
1
,
120
6
5
1
,
1
1
,
80
4
1
,-
40
3
2
.
1
0
1
,
5 (ppm)
0 6 (ppm)
Problem 102
- 1 ...........
1 ------- ' ' ’ • 1 ................ ..
'
1.2
80
60
40
Mass Spectrum
119 % of base peak
' 1 ' i 1 t t't ' 1 ' 1 ' 1 »1 ' 1 ' 1
100
1.6
2.0
-
_
w 0 CD 0
\ \
""
\ \
UV spectrum
M+* 134
\
2.4
20
lr
I „ C 10H 14 ........... 1-- •---1-- •-- 1-- ■-- 1-- •-- 1-- --- 1-- .-- 1-- .-- 1-- .-- 1-- 1_1_1_1_._1_1_1_1_1_._1_1_ 200 240 280 120 160 40 80
1
m/ e
200
250
300
350
X (nm)
n— ■ — i— 1— r 13C NMR Spectrum (125.0 MHz, CDCI3 solution)
DEPT
CH2i|f CH3f C H f
solvent
proton decoupled
i__ _i
i
1
i 1 200
i
1
i 1 160
i
1
i 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1------120 80 40 0 5 (ppm)
5 (ppm)
213
Problem 103
4000
3000
2000 1600 V (cm"1 )
1200
800
UV Spectrum ?tmax 260 nm (log10s ^10 2.5) solvent: methanol
40
80
120
160
200
240
280
m/ e 1
1
1
■.. "1
1
1
1
1
J
1
.
1
.
1
1
.
1
.
1
.
1
,
. . . . 1. 1 .
I
1
I
1
1
.
1
.
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
ch 2i)i ch 3|
CHf
solvent 1
proton decoupled
t
1
.
1
.
1
200
.
1
160
.
1
.
1
.
120
1
.
1
80
40
0
s (ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
r
7.5
7.0
ppm
r .......... >
_ A1 ,
10
214
8
.............- __________ ______
____
L
TMS
-
_I
1
0 8 (ppm)
Problem 104
4000
2000
3000
1200
1600 V (cm-1)
100
800
Mass Spectrum
147 139/141
80 60 40
M+‘ 182/184
20 C11H15C| 40
120
80
160
200
240
280
m/ e ........i..... 1 ....... i”.... " r
"i
1—
i
1---- T ..
r
1—
13C NMR Spectrum
1
1
'
1
1....... 1....
'
1
1
1
................
Resolves into two signals at higher field strength
(50.0 MHz, CDCI3 solution)
D EP T
'...... I
ch 2| ch 3| c h |
Resolves into two signals at higher field strength
proton decoupled
^
solvent
^
III I
.
I
.
I
200
10
9
.
I
1
1
160
8
7
*
I
*
l
1
120
6
5
l
*
I
4____ l____1-------1-------1-------1-------1----------------
80
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
215
Problem 105
4000
3000
1200
2000 1600 V (cm"1 )
800
UV Spectrum ^ max 262 nm (log10e ^10 2.5) solvent: methanol
40
120
80
200
160
240
280
m/e 1 1 1 ' ... . 1 13C NMR Spectrum
■".
1 ■1
1
J
T
'
J
u
>
i
1
1
t
1
1
1
1
1
'
1—
1:15 ppm
130 CH3|
r
expansion
(50.0 MHz, CDCI3 solution)
DEPT CH2|
•
T
CH f
i expansion
J
L
130
......... I
....... I
proton decoupled I * I ....
I
216
9
I
1
1
160
200
10
.i
_
8
7
1:*
PPm
1- J_.. »
1
.
1
1
1
.
80
120
6
11
^ solvent
JO
_.
5
4
.
40
3
2
1 0
1
8
(ppm)
0 8 (ppm)
Problem 106
4000
1200
2000
1600 V (cm‘1 )
3000
100
800
Mass Spectrum
183/185
UV Spectrum
80: ? lmax 258 nm (log1Qs 4.2) 60 155/157
solvent: ethanol
M+’ = 198/200
40
20 C8H 7 O B r 40
120
80
160
200
240
280
m/ e i
1
1
i
i
1
1
r
1
i
1
J" “ 1
T
1
|
|
I'
|
'
|
P
i
.
|
1
i
I.....
13C NMR Spectrum (100.0 MHz, CDCIj solution)
1 D EP T
CH2(| CH3f C H f
proton decoupled
solvent
i i
.
i
i
.
i
.
i
200
.
i
1 .
160
1
. i
i
i
.
120
i
.
i
j
80
— i
40
0
S(PPm)
i
i
1H NMR Spectrum
J
10
1
I
9
i
I
8
i
i
7
i
l
6
i
5
i
i
4
i
i
3
i
i
2
i
i
1
i
0 5 (ppm)
217
m/ e ' 1 1 “1— 1 ” 1 13C NMR Spectrum
1
1
X (nm) 1
1 T
1
1 .1..... 1
.
|
'
|
1
I i 40
I
i
I 0
i 5 (ppm)
expansion
(100.0 MHz, CDCIj solution)
D EPT
r
'.. 1 .1
CH2-j[ CH3t C H f
140 135
ppm
expansion
I I
proton decoupled
“r....... r------ r~
I •
I
i
solvent-r|n
............... LI
l i 200
I
i
I i 160
I
i
I i 120
i
I
^
I i 80
PPm I
i
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
___ | l r~
—
8.0
1---7.8
I L_ ___ « 7.6
T"" 7.4
1 ppm
—*
TMS
_ _ ..
L
---- 1
10
1
1
9
—1—
J—
8
1------
7
1
1 ------- 1 ------- 1------- 1------- 1------- !
6
5
4
1
3
1
1
2
1
1
1
1 1
»
1
0
6 (ppm)
218
Problem 108
-1 rTT» T»"T ' I * 1' 1' 1' 1' 1
1200
2000
1600 V (cm'1 )
3000
4000
100
40
20
base peak
60
Mass Spectrum
119
UV Spectrum
^max 254nm (log^c 4.3)
1
^max 258 nm (log^s 4.2 )
% of
80
800
^ max 279 nm (log1Qs 3.3)
M+' = 154/156
^ ~ | C gHyO CI — i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i . i . i . i . i . 120 160 200 240 280 80 40
A,max 290 nm (log10e 3.0) solvent: hexane
m/e I
1
'
'
I
'
1
I
1
I
■
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
c h 2|
CH3f
1
r
1
1
■ 1
'
1
•
1
•
1
.—
1—
.--------------
expansion
!
|
132
|
ch
131
ppm
|
expansion
I
I
; i 1 proton decoupled
132
. i
200
i
131
ppm
.
i
solvent 1
I
L
..................................... .........
( 1. A
.
i
160
............................................... .........
120
.
80
40
0
5
(ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution) expansion
—
_
I j — i—
10
1—
1—
9
1—
1—
8
r~
__
8.0
i I 1—
7
_^
j 7.5
7.0 ppm TMS
1 1
1
.
6
.
5
1
1
4
1
,
3
1
,
2
1
.
1
1
.
1
0 6 (ppm)
219
Problem 109
100
Mass Spectrum
UV Spectrum
80 ^max 224 nm (log10e 3.4) 60 s o l v e n t : m e th a n o l
40
20 C8H8^3 »
n. ■
i
■.. 1
1
1
1
1
13C N M R Spectrum (50.0 MHz, CDCI3 solution)
D EPT
c h 2|
CH3f
ch
|
solvent proton decoupled
|
1
li
-------1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1-- l_
200
160
120
i
1
i
1
80
i
1
i
40
1
0
i
S(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution) expansion with resolution enhancement
--------------- ».Ip---------------------- i ..... .....i.......
i
........... .
I i
160
I
I i
I
120
1
I
1
ill l
80
.
I
1
40
I
1H NMR Spectrum (200 MHz, CDCI3 solution)
s
i
TMS
1 j
10
8
L l_ 0
8 (ppm)
229
Problem 119
4000
3000
2000
1200
1600
800
V (cm-1) 135
100 cD 80 -■ ±C -. C D. Q 0/> 60 -■ < -. _C D Q 40 7 O
UV Spectrum A,max 262 nm (log10£ 2 .6 ) A,max 269 nm (log1Qs 2 .6 )
150
20 : :
Mass Spectrum
43
solvent: methanol
M + '= 192
i C i2 h 16o 2 ii , i. u ,i ___ _ I , i . i . i ,—i—.—i—.—i—,—i—,—i—.—i—,—i—,—i—,—i—,—i—.—i—.—i—,—i—,— 280 200 240 120 160 80 40
m/e I
1
1............... 1
1
1....... 1
1
1
1
|
.
|
1
|
|
1
1
|
1
1
1
1
1
l
*
0
8
13C NMR Spectrum MHz, CDCI3 solution)
(100.0
DEPT
CH2i|( CH3f CHf
1
expansion solvent
Iw ....r1.
proton decoupled
I
149 ppm
1
1 1 ----------- 1----- 1------1----- 1------1----- 1------1----- 1------1----- 1------1----- 1------1----- 1----- 1----- 1
1
I
200
10
230
9
160
8
7
120
6
5
l
1
l
80
4
»
l
40
3
2
1
(ppm)
0 8 (ppm)
Problem 120
4000
2000
3000
1200
1600 V (cm'1 )
800
UV Spectrum
A/max ~ 240 nm (l°g10s >4'°)
40
80
120
160
200
240
280
m/e 1
1
1
1
1
1
I
1
1
1 ...r
1
i " ' 1
1
i
1........ i ...... 1
i
..i ....... ■ "‘ i
1
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EPT
CH2| CH3f C H f
_ _ .....................................................
solvent
proton decoupled 1
i 1
1
.
1
.
■
200 1H
I
1
1
.
•
160
1________ i
.......................... .......... i 1
1
■
•
120
1
1
.
80
...-J____ ■----------
1
.
40
0
5(PPm)
expansior1
I
r----
8.0
I
1
1
■
NMR Spectrum
(400 MHz, CDCI3 solution)
1
1
1
7.Ej
PPM
TMS
I 1
i
1
1
1
1
1
1
1
1
--------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- __________ __________ 1 . .................
10
1
9
8
7
6
5
4
I
1
3
...............
i .........- ..... 1................... 1
2
.............
1
1
I
1
1
............ .................... .............
0 5 (ppm)
231
Problem 121
4000
3000
2000
1200
1600
800
V (cm‘1) UV Spectrum
X max 227 nm (log1Qs 3.9) ^max 305 nm (log^s 4.0) solvent: methanol
120
80
40
200
160
240
280
m/ e ....
r..... 1
!" '
'r
1
1
" ■
i
•
1
1
1
1— r
'
7
1
r
1 - ... 1
1
1
1
• ......
1
1
■
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
...... _
1
.
CH2i|f CH3f CH f
D EPT
proton decoupled
solvent
j
1
1
i
----------- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1------1----- 1----- 1----- 1----- 1----- 1----- 1----- 1
160
2 0 0
1
1
80
1 2 0
1
1
40
0
5 (ppm)
1H NMR Spectrum (400 MHz, CDCIj solution)
expansion
—
, 1___ 1 8.0
7.0
---
ppm
TMS
- i
1
10
232
1
1
9
1--
1
8
1---
7
1------- 1------- 1------- 1------- 1------- 1------- 1---
6
5
4
1
3
1
1------ 1
2
1
1
I
1 ------ 1
0 6 (ppm)
X(nm)
m/ e ■
i
1— r
1
1
1
1
'
1
“— 1
1
1 ... 1
I..... '
1
1
1
1
1
1
1
1................
,3C NMR Spectrurr1 (100 MHz, CDCI3 soluti on)
D EPT
ch 2| ch 3| c h |
solvent
* proton decoupled
.
1
1
1
1
1
200
10
9
1
1
1
1
160
8
7
1
1
1
1
1
120
6
5
1 1
1
______________ ______ 1 — 1----- 1----- 1----- 1----- 1----- 1----- 1-------------
80
4
40
3
2
0
1
S(PPm)
0 5 (ppm)
233
Problem 123
UV Spectrum
40
80
120
160
200
240
^m ax 225
nm (log^g8 4 . 2 )
^m ax 250
nm (log^c 3 . 8 )
^m ax 293
nm (log^s 3 .2 )
^m ax 330
nm (log^s 2 .6 ) solvent: methanol
280
m/e
1---- 1---- 1---- 1-------- 1— |----'— T
i
1
r
i
r
1
,3C NMR Spectrum (100 MHz, CDCI3 solution) expansion
i
i
134
D EPT
i 132
CH2ijf CH3t C H f
i
ppm
expansion
Ju l proton decoupled
-i
i
134
JL 132
ppm
X 1----1----1---- 1---- 1__ i----1__ i__ i---- 1----1__ i__ i__ i__ i__ i__ i__ t i i 200 160 120 80 40
i
t
i 0
i______ S(ppm)
1H NMR Spectrum (400 MHz, CDCL solution)
TMS
■i 11 1i 1" 1i 11 11
10 ppm
J
10
234
I
8
I
1______ 1______ i______ L
1
0 5 (ppm)
Problem 124
2000
1200
1600 V (cm'1 )
3000
4000
800
UV Spectrum ^max 2 77 nm (log1Qe 4 .2) solvent: methanol
120
80
40
160
200
240
280
m/e I
1
1
^
I
I
'
'
I
T~" | ■ 1 - |
1
|
1
|
'" I
.
I
■
1
1
...'
1
■
I
.
I
.....
13C NMR Spectrunn (100.0 MHz, CDCIj solijtion)
D EPT
c h 2|
CH3f C H f solvent
proton decoupled
1 i
I
I
■
I
•
I
200
■
I
•
I
I
■
160
I
.
I
.
120
80
I
40
0
S(PPrni)
expansion
1H Nl\/IR Spectrum (400 MhHz, CDCI3 solution)
__ ,1^ i
7.65
i
___;i[__ r ----r
7.45
i---- i
6.85
i —r
6.65 ppm
TMS
I ---- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1
10
9
8
7
6
5
4
1
i
3
1
i
2
1
i
1
1
i
1
0 5 (ppm)
235
Problem 125
UV Spectrum log10c > 4
40
120
80
160
200
240
280
m/ e 1— > — i— «— i— 1— i— ■ — r 13C NMR Spectrum
expansion
(50.0 MHz, CDCI3 solution)
DEPT
c h 2|
CH3f
ch
85
|
solvent
200
10
236
expansion
85
proton decoupled
160
8
120
80
80 ppm
80 ppm
40
0
§ (ppm)
0 5 (ppm)
Problem 126
UV Spectrum ^m ax 235
nm (log^s 4 . 1 )
A.max 2 6 5 nm (log10£ 3 .0 solvent:
40
80
120
160
200
240
)
ethanol
280
m/e
8 (ppm)
237
X (nm)
m/ e I
1
I
1
1
I
1
1 ..
1
|
1
1
1....
1
1
1
1
1
»
1
1 ..T
1
•
1
1
1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2{ CH3f C H f
solvent
proton decoupled
.
------ 1------- 1-------1------- 1------- 1------- 1-------1------- 1-------«------- 1-------•------- 1-------»------- 1-------1—
200
10
238
9
160
8
7
120
6
5
1
»
80
4
1
40
3
2
1
0
1
S(ppm)
0 6 (ppm)
Problem 128
100 80 60 40
20 40
80
120
160
200
240
280
m/e I
'
I
'
I
'
I
1
1
'
1
'
1
'
1
'
1
1
1
1
1
1
13C NMR Spectrum (100.0 MHz, CD3SOCD3 solution)
D EP T
CH2| CH3f C H f
1— ~1- solvent
proton decoupled
I
1
I
|
|
i
1
200
_j----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1------1----- 1----- 1----- 1------1------1-------------
160
120
80
40
1H NMR Spectrum
0
5 (ppm) expansion
(400 MHz,CDCI3/C D 3SOCD3 solution)
expansion
mi
________________________
in___
r solvent
11
10
9
8
7
6
5
4
1
. I___________ i________ __________ j i i i i i i i i__ 3 2 1 0 8 (ppm)
239
Problem 129
100 ;
120
Mass Spectrum
UV Spectrum
80 ^m ax
312 nm (log^s 4.3)
60 solvent: methanol
40
20 C8 H9 N O 40
80
120
200
160
240
280
m/e 1
I
1
1
I
1
l
1
I
i
■ "7
I
1
1
1
■
T1
1
'
1
1
1
1
,3C NMR Spectrum (100 Mz, DMSO-d 6 solution)
ch2\
D EPT
CH3f
ch
|
solvent — r proton decoupled 1
1
1
| 1
I 1
1
200
1
1
1
1
1
................... .. 1
120
160
n
1
1
1
1
80
1
1
1
1
40
1 0
1
5 (ppm)
1H NMR Spectrum (200 MHz, DMSO-d 6 solution) Exchanges with D2 O
I
residual H20 in solvent
TMS
1 10
240
8
1
0 6 (ppm)
Problem 130
1200
2000
3000
4000
1600 V (cm 1 )
800
UV Spectrum A,max 246 nm (log10e 4.2) ■'max 280 nm (log^s 3.1) solvent: methanol
40
80
120
160
200
240
280
m/ e 1
1
.
1
.
“1
’
'
1
1
T""
r
1
1
'
”1... 1
.1
1
1
T
i
—1
1
1
j— 1 —1------
1
1
,3C NMR Spectrum (100 MHz, CDCI3 solution)
D EP T
c h 2|
CH3f
ch
f
solvent
* proton decoupled
I
jj
1
— 1----- 1----- i----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1 —t-
200
10
9
160
8
7
1
i
120
6
i
»
i
80
5
4
40
3
2
0
1
S(PPm)
0 5 (ppm)
241
Problem 131
UV Spectrum
X max 250 nm (log10s 3.1) ^max 287 nm (log^s 2.2 ) solvent: chloroform
40
80
120
160
200
240
280
m/e I 1 1 '" I 13C NMR Spectrum
1.... 1
'
■7.....T— 7— 1...... r
1
'.... 1
1
1
1 ..1.... 1.......r "
1
1
1
1
1
(50.0 MHz, CDCI3 solution)
DEPT
c h 2|
CH3f
ch
|
solvent proton decoupled
I
i
i
_____________1 . ----------- 1----- «----- 1----- «----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- «----- i
200 1H NMR
160
80
120
40
1
1
1 0
8
(ppm)
Spectrum
(200 MHz, CDCI3 solution) expansion
/vi, n 7.6 exchanges with D20 on warming
10
242
1---- 1--7.0
ppm TMS
1 1
0 8 (ppm)
Problem 132
4000
2000
3000
1200
1600 V (cm'1 )
800
100
UV Spectrum
80 X max 250 nm (log10e 4.1) 60 ^max 285 nm (log^s 3.6)
40
solvent:
20 40
80
120
160
200
240
ethanol
280
m/e
200
10
9
160
8
7
120
6
5
80
4
40
3
2
0
1
S(ppm)
0 8 (ppm)
243
Problem 133
UV Spectrum ^ max 257 nm (log10e *10t 4.3) solvent: methanol
40
80
120
160
200
240
280
m/e ...!
'■
I
'
I
i
•
■
i
1
*
■ •—
r
1
1
.... r
r
' 1
...."r
t
1...
1
1
'
13C NMR Spectrum Resolves into two signals at higher field
(50.0 MHz, CDCI3 solution)
D EPT
ch
2| CH3f C H f
proton decoupled
1
.
I
.
.
Hi
1 I
.
I
.
160
200 1H NMR
solvent
i I I
Resolves into two signals at higher field
1
.
I
I
.
.
80
120
I
.
I
.
1
.
I
40
0
. 8
(ppm)
Spectrum
(200 MHz, CDCIj solution) expansions
V- pI [ p \ —i---- 1--- 1— 4.6
4.2
3.8 ppm
M
TMS
J
10
244
8
1
i
L
0 8 (ppm)
Problem 134
IR Spectrum
1730
(CHCI3 solution)
1200
2000
1600 V (cm'1 )
3000
4000
800
0.597 mg /1 0 mis path length : 0.2 cm solvent: ethanol
120
80
40
200
160
240
280
200
250
,
1
1
I
1
I
-1
I
300
350
X (nm)
m/e 1
1
1
1
1
1
n----- 1
I .1 80
II
I.
1
'
1
1
1
13C NMR Spectrum (125.0 MHz, CDCIj solution)
D EP T
CH2|
c h 3|
ch
|
solvent
proton decoupled
j
|
...... .-......... 1 >
10
I
I .I 200
•
I
I
9
I
8
. 1I L I. 160
7
I
I
6
I
.
I .I 120 I
5
I
I
I.
4
I
3
I .I 40
I
I
2
I
1
I
.
II 0
I.______ 5 (ppm)
0 5 (ppm)
245
X (nm)
m/ e
I I 200
10
246
9
I
8
I
I I 160
7
I
6
I
I I 120
5
I
I
I « 80
4
3
« ■ I i 40
2
I
1
«
I 0
»______ 5 (ppm)
0 5 (ppm)
X (nm)
m/ e I ' I ' 13C NMR Spectrum
I
I
'
'
I
1
1
1
1
,
1
|
1
’
.
1
|
|
1
(50.0 MHz, CDCI3 solution) Resolves into two signals at higher field
D EP T
CH2jf CH3t
ch
|
Resolves into two signals at higher field proton decoupled
|
ifj- solvent
^ .
1
1
«
1
1
200
«------- 1-------•-------1------- 1------- 1-------»------- 1-------•------- 1------- 1-------1-------1------- 1------- 1-------1-------1------- 1------- 1----------------
160
120
80
—I----- 1----- 1----- 1----- 1----- 1----- 1----- 1------ 1
i
10
5
9
8
Ill
7
6
I
i 4
40
I
i 3
i
' 2
i
' 1
0
6
(ppm)
i
i
i
0 6 (ppm)
247
X (nm)
m/e
expansion
13C NMR Spectrum CDCI3 solution)
(100.0 MHz,
II
D EPT 135
CH2| CH3f C H f
130
I h
1i |l
ji
'
126
ppm
125
expansion
ppm
note expansion
expansion
solvent 135
proton decoupled
130
ppm
126
125
ppm JL
_i
i
i
L_
200
160
120
80
40
0
5 (ppm)
1H NMR Spectrum expansion
(400 MHz, CDCI3 solution)
TMS
j\Rj 10
8
1
0 8 (ppm)
248
Problem 138
1200
2000
1600 V (cm'1 )
3000
4000
. 1. , . \ . , i , 1111111111 % of base peak
100
800
Mass Spectrum
UV Spectrum
75
80
? lmax 262 nm (log10s 2.3)
60
solvent: methanol
40
M+’ = 166 (< 1%)
135
91
20
» ......I,I ... c 10 h 14o 2 --------,— i— ,— i— ,— i— ,— i— ,— i— ,— i— ,— i— ,— i— .— i— ,— i— .— i— ,— i— ,— i— ,— i— .— i— .
40
120
80
160
200
240
I
1 " r
280
m/e I
1
I
1
I
'
I
1
I
1
1
r
~i_ t ..... r
'i
1
r
1
.
I
.
1
1............
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
ch 2| c h 3|
CHf
solvent
proton decoupled
I
.
I
.
200
10
9
in I
.
I
.
I
160
8
7
.
I
.
I
.
120
6
5
I
-
I
.
I
80
4
40
3
2
I
0
1
5(ppm)
0 8 (ppm)
249
Problem 139
1200
2000
1600 V (cm'1 )
3000
4000
800
UV Spectrum ? lmax 269 nm (log1Qs 2.7) X max 263 nm (log10e 2.7) solvent: methanol
m/ e --------------T”.. '
I
" '
l
'
l
1
'
1........1.......1...... T ......'........r
I
<
1
11
1
1
1
1............
I
.
I
.
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
proton coupled
..
.... .
.....
.......
i,
soh✓ent
proton decoupled I
,
I
| ,
I
| .
I
200 1H
.
I
__________ _____ ____ j I . I . I
.
160
120
.
I
_____ . I
80
.
40
0
5 (ppm)
NMR Spectrum
(200 MHz, CDCI3 solution)
expansion
1
1
2.5
2.0
1
1
1.5 ppm
TMS
__ . ... J I
1 ,, 1
10
250
1
1
9
1
1
8
1
1
7
•
1
6
1-
5
1 -
1---------1----------1--------- 1----------1----------1----------1----------1----------1----------1-----
4
3
2
1
0 8 (ppm)
Problem 140
! \
UV spectrum 0.491 mg /1 0 mis path length : 0.2 cm solvent: ethanol
40
120
80
160
200
240
200
280
250
1
'
1
1
'
1
1
300
350
X(nm)
m/e 1
1
1
■
1
1
1
1
1
r
1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
CH2(f CH3f
ch|
solvent proton decoupled
I
1 --- 1--- 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1_ .J_ _j___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1_______
200
160
120
40
80
0
5 (ppm)
1H NMR Spectrum (400 MHz, CDCIj solution)
expansion
expansion
-i-------- 1-------- 1— 4.6 4.4 ppm
_l
1
1----------
1.6
1.4
ppm
TMS
-i 10
1----- 1----- 1----- 1----- 1----- 1___ I___ i___ I___ i___ I___ i___ I___ i___ I___ i 8
I
■
1
0 5 (ppm)
251
X (nm)
m/e 1 -~r
I
-1
1
1
1
1
1 —.
.
|
|
-r
1
|
|
,
1
1
|
,
,
1
1
1
1
1
1
1
1
1
13C NMR Spectrum ( 100.0 MHz, CDCI3 solution)
D EPT
c h 2|
CH3f C H f
solvent proton decoupled
1 ----- 1------1----- •------1----- 1------1----- 1----- 1----- 1------1----- 1------1----- 1----- 1----- 1----- 1
200
160
120
1
80
40
0
5 (ppm)
1H NMR Spectrum (400 MHz, C D C I3 solution) expansion
expansion
expansion
—1------ 1
7.8
7.6
4.4
ppm
4.2 ppm
1.4
r
1.2 ppmi TMS
__ L
10
252
9
8
7
6
5
4
3
2
1
0 5 (ppm)
Problem 142
100 80 60 40
20 40
80
120
160
200
240
280
m/e --------- 1........1
I
1
I
1
I
1
I
—1
1
'
T - '
1
(50.0 MHz, CDCI3 solution) 135
D EPT
T"
■
I
1
I
1
I
1
1
1
I
1
I
< I
1
'
expansion
13C NMR Spectrum
1
i
130
PPm
CH2(f CH3f C H f expansion
!L
11
135
... . 1
1
130
ppm
solvent proton decoupled
..- ....... L_ j
|
I ... .
I
200
10
9
1
1
.1
I
160
8
11
A
7
1
1.
1
I
<
120
6
5
I
<
80
4
40
3
2
0
1
8 (ppm)
0 8 (ppm)
253
Problem 143
UV Spectrum
^max 220 nm (log^c 3.7) ^max 274 nm (log10s 3.3 ) solvent: ethanol
m/e
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
c h 2'|(
CH3f
ch
|
solvent
I
proton decoupled
j
1
i . 200
i
1
i 1 160
i
1
i , 120
i
■ 1 80
.
i
-I
I
40
L.
J
L
1
0
8
(ppm)
8 (ppm)
254
Problem 144
1600 V (cm'1 )
100 80 60 40
20
1200
2000
3000
4000
163
800
Mass Spectrum
UV Spectrum
- ^ ■ CD . Cl
? lmax 223 nm (log1Qs 3.9)
■ C/5 _ . -Q
^max 275 nm (log^s 3.1 )
0 0 0
“
‘o
:
^
[
I
:
it
i i
.
i
i
i
.
i
i .
» i
> i
.
I.
i i
+ '=
1 9 4
i
i
X max 281 nm (log10s 3.0 )
c 10h 10o 4
I.
i
120
80
40
M
135
1 0 4
.
i
.
.
.
200
160
a
.
i
i
solvent: ethanol
i
280
240
m/e ....1—
1
1
1 ......1....” T"
1
1
1
1
111
1
!...... r....... 1...... T
1 ... 1
1 .... r ".. 1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
CH2iji CH3f C H t
solvent proton decoupled
*
i
1 I
.
!
.
I
200
10
9
.
I
.
I
160
8
7
.
I
.
I
.
120
6
5
I
.
I
1
1
80
4
.1.. -
40
3
2
1
-L----- 1----- i------------
0
1
8 (ppm)
0 8 (ppm)
255
Problem 145
4000
3000
2000 1600 V (crrf1 )
1200
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
I
I
I
m/e
1
1
I
1
T
I
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
c h 2|
CH3f C H f
solvent
proton decoupled
I I__ J
200
160
1
L
120
80
40
0
8
(ppm)
8 (ppm)
256
Problem 146
100 : -
60
-
CD CO CO o
7
o
800
Mass Spectrum
91
80
40
1200
1600 2000 V (cm'1)
3000
4000
UV Spectrum
CO CD
^m ax
261 nm (log1Qe 3.5)
M + '= 92
solvent: ethanol
20
c 7h 8 120
80
40
200
160
240
280
m/e I
1
I
1
I
1
I
1
I
T"
1
1
1
1
1
1
1
1
1
1
1
1
1
'
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2(( CH3f C H f
solvent proton decoupled
I
.
I
.
I
200
10
9
.
I
.
I
160
8
7
.................... I « I_ .
120
6
5
t i I
. ..J___-•___ I------.------1----- 1------1----- *------------
80
4
40
3
2
0
1
5
(ppm)
0 6 (ppm)
257
Problem 147
4000
3000
1200
2000 1600 V (cm’1 )
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
I
1
I
1
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2ijf CH3t
ch
|
proton decoupled solvent i
I
.
I
.
I
.
I
200
J
160
120
i
L
80
40
0
8 (ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
J
10
258
i
I
9
i
I
8
i
I
7
i
I
6
i
5
I
i
4
I
i
3
L
2
1
0 8 (ppm)
1200
2000
1600 V (cm'1 )
3000
4000
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
i
i
m/ e i
r
r
•
i
1
i
•
i
1
r .. 1
i
>
i
i
«
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2i|( CH3f C H f
solvent
1 proton decoupled 1
■
1
^ -
1
200
10
9
^ .
1
•
1
160
8
7
•
1
«
1
«
120
6
5
1
«
1
.
1
80
4
«
1
40
3
2
-i
-L.... .
0
1
8
(ppm)
0 8 (ppm)
259
Problem 149
4000
3000
40
2000 1600 V (cm'1 )
120
80
160
200
1200
240
800
280
m/e i
1
i
1
1
r
1
1
f"
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2\f CH3f C h {
proton decoupled
solvent
I
8 (ppm)
260
Problem 150
100
. M + '= 128
80 -' 60
. -■ _
40
20
. 7 :
Mass Spectrum
03 CD CL
No significant UV
CD C/3
03
-Q 0 ^8 0
absorption above 220 nm 55
83 ,
100
1
C 7H 12O 2 , 113 i ---- 1— 1— I— 1— I— 1— I— 1— I— •— I— 1— I— 1__I_____1__1__1_____1 . 1 . 1 . 1 . 1 . 1 .
40
80
120
160
200
240
280
m/ e l
1
r
1
i
1
r
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
c h 2|
CH3f C H f
solvent proton decoupled
5 (ppm)
261
Problem 151
4000
2000
3000
1200
1600 V (cm'1 )
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
i — 1— i— «
i
1
i
1
i
1
i
1
"1
r
'
1
1
13C NMR Spectrum
1
*“
expansion
(100.0 MHz, CDCI3 solution)
DEPT
c h 2J
CH3f
ch
1-------- 1------- r 28 27 ppm
|
expansion solvent
proton decoupled
1-------- 1------- 128 27 ppm
I - I
J
1
I . 200
I
1
I 1 160
I
1
I 1 120
I
1
I 80
....
1
I
.
I 40
1
I
1
I 0
1_____
5 (ppm)
5 (ppm)
262
Problem 152
1200
2000
1600 V (cm"1 )
3000
4000
800
No significant UV absorption above 220 nm
40
120
80
160
200
240
280
m/e I
1
1
1
1
■1— r .... r "
1
1
•“
1
■
1
1
'
1
1
1
1
1
'
1
1
,
1
!
0
5 (ppm)
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
CH 2| CH3t
ch
|
solvent proton decoupled
_____ L... 1
1
1......j__... 1,
200
10
9
1
,
1
,
160
8
7
—
1----- 1— 1
120
6
j— 1
1
80
5
4
1
.
40
3
2
1
0 5 (ppm)
263
Problem 153
100
:
80
-
■ -
03 CD
60
~ • -
CD CO 03 o
7
o
40
1200
2000
1600 V (cm'1 )
3000
4000
800
Mass Spectrum
30
No significant UV absorption above 220 nm 56
“
20
M + ‘ = 1 0 2 (< 1i
L
__ i___ i___ i___ _
>iJi i
i i, •
1% )
C 5 H 14N 2
,1
................................................
i ___ 1___ 1___ ____ 1___ .___ 1___ i----- 1----- 1----- 1----- .----- L.
120
80
40
200
160
240
,
.
280
m/ e I
1
1
I
i
■
i
■
I
1
1
'
1
1
1
1
1
1
•
1
1
■
1
13C NMR Spectrum r 'in n
DEPT
c h 2{
CH3f
ch
|
solvent if
proton decoupled
1 I
1
I
1
I
200
10
264
9
1
I
1
160
8
7
1
I...._ J ___1
1
1
120
6
5
1
1
..... 1....... 1........L ..... 1 ,, 1........1 ...... 1,.... JL.................
80
4
40
3
2
0
1
8 (ppm)
0 8 (ppm)
Problem 154
100 :
Mass Spectrum
31
80 ’ CO
: o-
No significant UV
0j) 60 ■< 40
~-Q 7o -^
absorption above 220 nm
M+' 100 < 1%
20
c 2h 3 f 3 o
^ IJ . iJ ... i _____i_i_i_i_i_*—
i— *— • * » • i — i— . i_j_1_1_1_1_1_1_1_._1_._1_._ 280 120 160 200 240 80
40
m/e ...1........1
1
1
1
" 1
I
1
1
J
"I
1
I
1
"1........r
1
1
1
1...... r
1
-l
T
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
— —
ch2| CH3f chf 1 1
V
I |
v
61.5
expansion
1 60.5 ppm
expansion l
solvent
proton decoupled
I
!i
— i------1------1— 61.5
J
........................................................................................I________I________ I________ I________ I________ L_
200
10
160
8
120
60.5 ppm
80
40
L-
0
1
8
(ppm)
0 8 (ppm)
265
Problem 155
100
:
80
-
60 40
119
■ “. -Q -
CO CD
-
CD CO CO
Mass Spectrum
UV Spectrum
91
log10s > 5
M+‘ 226
7 0
20
| , I C 15H 1 4 ° 2 I „ l —--L-.-- .--1--.--1--.--1--.--1--.-- --.-- --.-- --.--1--.--1--.--1--.-- --.-- --.-- --.-L
1
40
80
120
1
160
1
1
200
1
240
1
280
m/e
i
1
r
13C NMR Spectrum expansion
(150.0 MHz, CDCI3 solution)
DEPT
CH2| CHat C H f
130
129
128 ppm
expansion
130 proton decoupled
129
J
200
128 ppm
_____________solvent -rj~i
160
120
I
I
80
I
I
I
I
40
L.
J
L-
0
5 (ppm)
8 (ppm)
266
Problem 156
1200
2000
1600 V (cm'1 )
3000
4000
100 :
105
800
Mass Spectrum
80 -• 00 - CD
log10e > 5
. Q.
60 -
UV Spectrum
0
(/)
- CO . _Q
M+’
40 • ° -^
226
77
20
121
1
n
U
I I C 15H 1 4 ° 2 .i... i 1]__iLJ I s.l ------- .— i—.— i—.— i—.—j _i_i_ --- 1------- 1---.--- 1---1--- 1 _1_1_1_1__1_1_1__ » - » ■ ! . 40
80
120
160
200
240
280
m/e
200
10
9
160
8
7
120
6
5
80
4
40
3
2
0
1
5 (ppm)
0 5 (ppm)
267
40
80
120
160
200
240
280
200
250
I
1
1.
1
1
1
1
1
300
350
A,(nm)
m/ e 1 —r
1
1
|
1
|
1
|
1
1
1
I
'
1
*
I
1
1
1
expansion
13C NMR Spectrum (75.0 MHz, CDCI3 solution)
DEPT
135
CH2^ CH3f C H f
125
ppm
expansion solvent
-r-1 il 1 T 1 --- r- T---- T.. —
proton decoupled
135
...i_____________ L______ i ----- 1----- 1----- 1------1----- 1----- 1----- 1----- 1-----j_----- 1------1------1----- 1
200
10
268
9
160
8
7
l
1
120
6
5
l
i
125
t
1
1
80
4
ppm
1
40
3
2
0
1
5 (ppm)
0 5 (ppm)
40
80
120
160
200
240
280
200
250
T
'
I
'
I
1
I
1
300
350
A (nm)
m/ e I
'
1
1
r.....
1
i
1
r
1..
i
.nr"
I
•
i .. 1
1
r
expansion
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EP T
... f
CH2-If CH3f C H f
140
13-0
— T
ppnn
exparision solvent 11
— t
T........ 1 -.. T
proton decoupled
I
.
I
i
*
l
200
10
9
i
I
i
I
•
160
8
7
13 0
140
i
I
i
l
pprn
>
120
6
5
I
•
l
•
80
4
I
40
3
2
•
l
0
1
>
5 (ppm)
0 5 (ppm)
269
Problem 159
2000
1200
1600 V (cm'1 )
3000
4000
800
UV Spectrum l°9io e > 4
40
80
120
160
200
240
280
m/e
1----1 1
1
I
1
I
r
n
1
i
1
i
1
i
-t
n
|
,
expansion
13C NMR Spectrum
r
expansion
(100.0 MHz, CDCIj solution)
D EPT
CH2j CHst
ch
|
128.7 128.4 ppm
130
expansion
120
ppm
expansion • solvent
proton decoupled 130
128.7 128.4 ppm
1----- 1------1----- 1----- 1----- 1------1----- 1----- 1----- 1___ I___ i___ I___ i
-i
200
10
270
9
160
8
7
120
6
5
I
i
I
80
4
■
I
40
3
2
i
120
I
0
1
ppm
i
S(ppm)
0 5 (ppm)
Problem 160
UV Spectrum log10s > 4
80
120
160
200
240
280
m/e
5 (ppm)
271
Problem 161
4000
3000
2000 1600 V (cm'1 )
40
80
120
160
1200
200
800
240
280
m/e I
1
...1
1
1
1
CH3f
ch
1
1
1
1 ........r "....1
'
'
1
1
1
' ......T ” ....'
1
'
1
1
1
1
1
1
------
13C N M R S p e c tru m (50.0 MHz, CDCI3 solution)
DEPT
c h 2|
|
proton decoupled
solvent
| 1
it
----1-------•-------1-------1-------1-------•-------1-------1------1-------1-------1-------1-------1------J------ 1
200
160
120
L
1
80
1
40
0
1
5 (ppm)
1H NMR Spectrum (200 MHz, CDCIj solution)
expansion
expansion scale
8.0
8.5
10
272
9
8
7
6
5
4
ppm
3
2
1
0 S (ppm)
Problem 162
1200
2000
1600 V (cm'1 )
3000
4000
100 ;
800
Mass Spectrum
119
UV Spectrum
80 ^m ax
246 nm (log^s 3.7)
60 ^max 273 nm (log^c 2.6 ) 40
Mh
20
solvent: ethanol
191
c 12h 17n o 40
80
120
200
160
280
240
m/e
200
10
9
160
8
7
120
6
5
80
4
40
3
2
0
1
5 (ppm)
0 6 (ppm)
273
Problem 163
1200
2000
1600 V (cm'1 )
3000
4000
800
UV Spectrum
I
max
277 nm (log^s 3.4) solvent: methanol
120
80
40
200
160
240
280
m/e 1...
1
I
1
I
'
1
'
1
1
1 .... 1...
1
1
1
•
1
1
1
1
1
r
1
1
1
1
1
1
1
1
»..
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
c h 2j c h 3| c h |
solvent proton decoupled
|
in
--------------1-------1-------1-------1-------1-------1-------1-------•------- 1-------»------- 1-------»------ 1
200
10
9
160
8
7
j—
120
6
5
1
80
4
40
3
2
0
1
5 (ppm)
0
8 (PPm)
274
X (nm)
m/e
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
CH2| CH3f
ch|
solvent
expansion expa
jl
T------------ 1-------------T 175 170 165 ppm
1
proton decoupled I
»
I
I
I
200
10
9
.
I
J
L.
160
8
7
120
6
5
i
I
i
I
80
4
i
L
40
3
2
0
1
8 (ppm)
0 8 (ppm)
275
A,(nm)
m/e --------
I
'
I
'
I
'
I
1
T
1
i
1
|
■
|
.
|
,
|
.
|
i
|
.
i
1
i
1
i
1
i
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f C H f
r
solvent
proton decoupled
----- 1------1----- 1------1----- 1----- 1----- •----- 1----- 1------1----- «----- 1----- 1----- L
200
10
276
9
160
8
7
i
120
6
5
1
i
1
80
4
40
3
2
0
1
5 (ppm)
0 6 (ppm)
m/e I
.
I
.
I
.
I
■
X(nm) 1
“I
1
I
i
' 1
1
1
1
I
i
.
i
T
1
1 T
.
i
.
..... 1
13C NMR Spectrum (150.0 WIHz, CDCIj solution)
D EP T
CH2| CH3f C H f
solvent proton decoupled
I !
.
i
i
.
i
.
200
.
i
1
I .
i
.
160
i
.
i
.
120
i
1 .
80
40
i
0
.
S(ppm)
1H NMR Spectrum (600 MHz, CDCI3 solution)
Expansion Scale
exchanges with D20
7.00
6.98
6.72
6.70
6.68 ppm
vertical expansion x20
TMS
_L J ---------------- 1---------------- 1---------------- 1---------------- 1---------------- 1___________I__________ I___________I__________ I__________ I__________ I__________ I__________ I__________ I__________ L_
10
9
8
7
6
5
4
3
2
1
0
5 (ppm)
277
Problem 167
100
1200
2000
1600 V (cm-1)
3000
4000
■ -
60
* -
-
Mass Spectrum
261/263/265
:
80
800
182/184 CO CD 0 C/5 CO o
M+*
197/199
40 7
276/278/280
o _ -o
20
C9H10 Br2 _l
40
i
I i
80
I
I
L_
120
_l
160
i
I
200
.
I
.
I
I
240
■
t
I
u
L_
280
m/e I
1
I
'
I
'
1
'
I
'1
1
T_
1
1
T .. '
1
'
1
1
1
1
1
■
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
L D EP T
c h 2|
c h 3|
ch
1
|
solvent proton decoupled
i
.............111 .
______ 5 (ppm)
________ I____ 1____ 1____ 1____ 1____ 1____1____ 1-------1-------1-------1-------1---------------1-------1-------.-------1-------1-------.-------.-------.____ .____ 1____ 1
200
10
278
9
160
8
7
120
6
5
80
4
40
3
2
0
1
0 6 (ppm)
Problem 168
i ■» . ■■■• ■- I - i ■« ■t • • ^*
4000
3000
i— i
2000
i
i
t
— i— I— i— i— •— i—
1600
1200
800
V (cm'1) 100 80 60 40
20
m/ e i
i
'
1
1
r
r
I
(50.0 MHz, CDCI3 solution)
DEPT
'
one signal even at high field
13C NMR Spectrum
c h 2| CH3f c h |
Resolves into two signals at higher field
solvent proton decoupled
1____ 200
10
9
J_ _ L
J
160
8
7
6
5
4
I
I
3
2
I
I
0
40
80
120
1
L.
5 (ppm)
0 6 (ppm)
279
m/e I
I
I
I
I
1
I
1
A,(nm) 1
I
1
1
1
1
1
1
1
'
1
1
1
1
,
1
.
1
1
'
13C NMR Spectrum (150.0 MHz, CDCI3 solution)
DEPT
ch 2| c H3f C H f
solvent proton decoupled
-
-L
1
■
1
1
-L
200
■
1
1
1
160
■
.....................................................1
120
80
40
1H NMR Spectrum
.
0
5 (ppm)
expansions
(600 MHz, CDCI3 solution)
TMS
Expansion Scale 0
10
280
9
8
7
6
5
4
3
5
10 15 20 Hz
2
1
0 5 (ppm)
Problem 170
4000
3000
2000
1600
1200
800
V (cm'1)
40
80
120
160
200
240
280
m/e
1
.
1
1
1
1
'
1
....1
1
1
'
1
1
1
1
1
1
1
1
'
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
CH2(f CH3f C H f
proton decoupled
1
1
......... ....... .............. . 1 . 1 .
200
......I ..... J 1 . 1
160
I .
solvent
L----1 .
.
.. 1
.
120
. 1
1
______ . 1
____ 1 .
80
1
40
.
, ...
1
0
8 (ppm)
1H NMR Spectrum (2 0 0 M H z , C D C I 3 so lutio n)
R e s o lv e s into tw o sig nals of e q u a l intensity a t h ig h e r field
7 .0
7 .5
p pm
TM S
_______«UUL _1_________ I_________ I_________ I_________ L .
10
9
8
7
6
5
4
3
2
1
0 8 (ppm)
281
Problem 171
3000
4000
1600
2000
1200
800
V (cm‘1)
40
80
120
160
200
240
280
m/e
200
10
282
160
8
120
80
40
0
6 (ppm)
0 6 (ppm)
Problem 172
4000
3000
2000 1600 V (cm"1 )
1200
800
m/ e i
1
i...~“r ..T "
i....r
*
1
T " T ....T...... r
1 ... i
1
i..
1
i
1
1
,
i
1.............
1
i
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT CH2 |
CH3|
CH f solvent
proton decoupled
1
1
.
1, I
.
200
.
I
1
II,
.....................................1
160
1 , 1
,
120
1
,
,............
80
40
0
...
5 (ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
1 10
9
8
7
6
5
4
3
2
1
0 8 (ppm)
283
m/e
200
X (n m )
160
120
80
40
0
8 (ppm)
1H NMR Spectrum (100 MHz, CDCIj solution) 6H
TMS
-IH
residual CHCI3 insolvent
1H
JL 10
284
9
8
7
6
5
4
3
2
1
0 8 (ppm)
Problem 174
100
68
Mass Spectrum
UV Spectrum
80
A,max 225 nm (log10s 3.9)
60
M+ '=
40
s o lv e n t:
96
methanol
20 c 6 h 8o _J
I
I
I
40
I
I
I
1
I
I
120
80
I
I
I
I
I
I
I
.
I
■
200
160
I
*
I
.
240
I
.
I
280
m/e
I 1 I 1 13C NMR Spectrum
I
1
I
1
1
1
|
1
|
1
|
1
|
1
|
1
,
,
,
.
( 100.0 M H z, C D C I3 solution)
DEPT CH2| CH3f CHf
1
proton decoupled
______ I
»
solvent
i I
200
1
1
1
* ................. 1 . ____ 1 *___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1-----1-----1-----1-----1-----1-----1---- 1---------0 8 (ppm) 40 80 120 160
8 (ppm)
285
Problem 175
2000
3000
4000
1200
1600
800
V (cm‘1 ) UV Spectrum ^max 263 nm (log^s 3.9) X max 305 nm (log10e 2.1 ) solvent: hexane
40
80
120
160
200
240
280
m/e ■............... I”
1
1
r .. r ™...i
..................................................
13C NMR Spectrum
1_ .....r — «"
expansion 40
c h 2|
CH3f
ch
1
"1.......1
1 ....
1
1
1
i
(100.0 MHz, CDCI3 solution)
DEPT
1
i i t—i—r 30
f
T—1--1--1--1— 20 ppm
| expansion
i —i—i
-r 30
40
proton decoupled
20 ppm
solve
nt 1. •
»
1
200
»
1
»
1
160
>----- 1---- ‘
i
120
»
>
'
l
80
i____ i
l
i
l
40
i
I
i
I
0
i
6 (ppm)
5 (ppm)
286
Problem 176
UV Spectrum ^max 234 nm (log^c 4.1) solvent: ethanol
40
80
120
160
200
240
280
m/ e
6 (ppm)
287
Mnm)
m/e '
I
1 “T
1
n— 1
1
1
1
r
1
’
1
'
1
1
1
'
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
l3C NMR Spectrum (100 MHz, CDCI3 solution)
D EPT
c h 2|
c h 3|
ch
|
solvent proton decoupled
*
--- 1----- 1----- 1------1----- 1----- 1----- 1------1----- 1----- 1----- 1------1----- L-----L
200
10
288
9
160
8
7
1
120
6
1 1
80
5
4
40
3
2
0
1
8 (ppm)
0 8 (ppm)
Problem 178
4000
2000
3000
1200
1600
800
V (cm'1) 100
UV Spectrum
80
^"max 260 nm (log^s 2.9 )
60
^max 266 nm (log^s 3.1)
40
X max 272 nm (log10s 3.1 )
20
solvent: methanol
40
80
120
160
200
240
280
m/e I
.
1
I
,
.
I
.
I
.
I
■
1
I
I
1
I
.
I
'
I
1
I
1
I
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
c h 2| c h 3! C H f
solvent proton decoupled
I
I I
.
I
■
I
200
10
9
•
I
.
«
I
160
8
7
.
I
.
I
.
120
6
5
I
.....i..
____ I____ .____ I------- .--------------
40
0
80
4
3
2
1
8 (ppm)
0 8 (ppm)
289
Problem 179
1600 2000 V (cm1)
3000
4000
1200
800
100
UV Spectrum
80 ^m ax
=
243 nm (Iocj^qS 4.1)
60 ^max = 291 nm (log1Qs 3.4) 40 solvent: ethanol
20 120
80
40
200
160
280
240
m/ e t—
■ — r
expansion
13C NMR Spectrum (100.0 MHz, CDCIg solution) 128
DEPT
CH2j CH3f CHf
126
124
ppm
124
ppm
expansion
128
126
proton decoupled solvent
J
1
I
I
1
200
10
290
9
1
I
1
I
160
8
7
.
I
1
I
1
120
6
5
I
I
1
I
I
1
80
4
I
1
40
3
2
.
L
0
1
5 (ppm)
0 8 (ppm)
"i « rm f r ' - ... 1-- ■ -- |-- --- i-- --- |-- ■ -- r » |--
| 1i ' » ' i ' i ' | '
IR S p e c tr u m
r
Problem 180
1751
(CHCI3 solution)
i • » * ■ • ■• » • I • ■ - ■ • ■ ■- I »_»___ 1.i * 1 ____1____L 800 1200 1600 2000 3000 V (crrf1)
4000
UV Spectrum
X max 268 nm (log10s 3.1) A-max 275 nm (log1Qs 3.1) s o lv e n t: ethanol
40
80
120
160
200
240
280
m/e i
'
i
1...“ i
1
i
-T—
r — r'
1
I
1
I
I
1
I
.
______ I .
.
I
1
I
1
I
.
I
.
I
.
I
..X.
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2| CH3f C H f
proton decoupled i I
I
.
,
solvent ..................................... I_____ , I , I . I .
200
10
9
160
8
7
1 I
.
I
.
120
6
5
I
80
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
291
Problem 181
4000
3000
2000
1200
1600
800
V (cm'1) UV Spectrum ^max 249 nm (log10s 4.1) ^ max 292 nm (log^s 3.3) solvent: ethanol
40
80
120
160
200
240
280
m/e
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
c h 2| CH3t c h |
ppm
130
125
expansion
proton decoupled solvent ppm
130
125
_L
200
160
120
J 80
I 40
I 0
8 (ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
] 10
292
8
lui.
TMS
jj 1
0 8 (ppm)
Problem 182
UV Spectrum ^max 265 nm (log^c 3.0) ^max 272 nm (log^s 3.1) X-max 294 nm (log10e 1.9) solvent: ethanol
40
80
120
160
200
240
280
m/ e
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
expansion
expansion
IT 3.2
T---------- 1— 7.3 7.2
3.0
ppm
,
1
1--
2.6
2.5
ppm
TMS ppm
_JL J
10
1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1_____ 1_____ 1_____ I_____ I_____ I_____ I_____ I_____ I_____ I_____ I_____ L.
8
1
0 5 (ppm)
293
Problem 183
1200
1600 2000 V (cm-1)
3000
4000
100
800
Mass Spectrum
80 -■ 03 -. 0C 60 -■ C0/> - 0 . £ 40 7 0 - 0 ^^
^m ax
M+ '= 180
l
UV Spectrum
1£32
^max 330 nm (log^s 4.2) solvent: ethanol
20 »
265 nm (log^s 4.5 )
*
»
*—
1
— *— »—
•— •— •— •— .—
40
80
1
1
ini
— •— •— •— •—
120
•—
160
I 1
— •—
1 —
•—
200
1 —
*—
1 —
240
c 13h 8 o •—
1 —
•—
1 —
•—
1
— •—
280
m/e
i— ■ — i— •— i— 1— i— 1— r 13C NMR Spectrum
expansion
(100.0 MHz, CDCI3 solution)
135
DEPT
125
ppm
CH2^ CH3f C H f -------------expansion
solvent
JL
proton decoupled
i
135
—1------ r-
125
I 1__
ppm
_J__________________ L 2 0 0
160
120
80
40
0
5 (ppm)
8 (ppm)
294
Problem 184
100
Mass Spectrum
45 43
80
29
No significant UV
60
absorption above 220 nm
40
M+’ = 132 (< 1%) 89
20
87 |
117 131
c 6 h 12o 3 ,1
40
80
120
160
200
i
I
I
240
I
I
L_
280
m/e
1— 1— i— 1— r
“i
1
11— r•— r
I
I
I
I
I
I
I
J
I
I
I
[-
I
1
I
13C NMR Spectrum (150.0 MHz, CDCI3 solution)
D EP T
ch 2J c h 3|
CHf
solvent proton decoupled
I
.
I
200
.
I
.
I
160
I
I
I
I
120
I
I
L_
80
40
I
L
0
5 (ppm)
6 (ppm)
295
Problem 185
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e 1
..I
I
'
1
I
I
'
1
I — 1
1
1
^
1
1
1
1
1
'
I
'
I
'
I
i
I 0
i 5 (ppm)
13C NMR Spectrum CDCIj solution)
(100.0 MHz,
D EPT
CH2| CH3f CHf
solvent proton decoupled
i
I
i
I i 200
| I
i
I i 160
I
i
I i 120
I
i
I
I i 80
I
i
I i 40
1H NMR Spectrum (400 MHz, CDCI3 solution)
■ ----
__ _
__
TMS 1
1
10
1
1
1
1
8
1
i
i
1
*
1
1
1
1
I
.
1
1
1
1
1
I
0
5 (ppm)
296
Problem 186
1200
2000
1600 V (cm'1 )
3000
4000
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e I
.
I
.
1
I
I
1
I
1
1
'
1
'
1
'
“ 1-----T“
'[
1
1
1
1
1—
4.0
40
80
120
160
200
240
280
m/e I
1
1
' .....1
■
1
'
1
1
T
1
r
...... r.... '
t .... ' 11' "i.....
1
1 1 1
1
1....
1 0
1 8 (ppm )
13C NMR Spectrum (50.0 MHz, CDCI3 solution) two resolved signals at higher field
D EPT
CH2(f CH3f
ch|
expansion
—
two resolved signals at higher field
i
iL
solvent
iii
1
1
.....
8,0
70 ppm
i
proton decoupled
----------- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- «----- 1----- 1----- 1----- 1----- 1----- 1----- 1---- 1------ 1----- 1
200
10
304
9
120
160
8
7
6
80
5
4
1
40
3
2
1
1
0 8 (ppm)
Problem 194
1200
2000
1600 V (cm'1 )
3000
4000
800
100
UV Spectrum
80
lo g 10e > 4.0
60 40
20 120
80
40
160
200
240
280
m/e I
■
I
1
I
■
I
1
I
■»
i
1
• ..i
1
i
i
1
1
'1
'
1
1
'
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
ch 2J
CH3f
ch|
r solvent
proton decoupled ..... l
_____ ______i —... L ._ ....... i i i i i < i i
.
200
10
9
160
8
7
i i
i
i
.
i
120
6
5
.
i
i
i .....i..... j ___ .___ i— .------------
80
4
40
3
2
0
1
5(ppm)
0 6 (ppm)
305
Problem 195
1600 V (cm'1 )
100 : 80 60 40
1200
2000
3000
4000
Mass Spectrum
121/123
•
03
7
Cl
800
41
No significant UV
” 4.0)
M+*
60
227/229
40
197/199
20 C 8 H6 N 0 2 Br 40
80
120
160
200
240
280
m/ e i— 1— r 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
CH2j CH3f
ch|
solvent
Resolves into two signals at higher field
proton decoupled
200
160
120
80
40
0
5 (ppm)
6 (ppm)
327
Problem 217
100 :
91
Mass Spectrum
107
80 • 03 -. 0 Q. 60 -• 0 0 -. -Q 03 40 7 O M + '= 166
20 ! * • *
I, CmHmO? I III,I li .1 j! li I 1,I ............ *—* J-- ■ J-- - -J-- •-- 1-- *-- 1-- •-- 1-- ■-- 1-- CD
.
-Q
7
o
absorption above 220 nm
-
20
102
j*
-I
I I
llll IM il il I I
c 5H9N 0 4
M + ' = 1 4 7 « 1 %)
lL_
I I—
_J
I
I
I____I— I___I
120
80
40
I
I
I
I
I
I
I
200
160
I
I
I
■
I
280
240
m/e i
'
i
r
i
T
r
•
i
1
i
•
1
i
i
•
i
..... I” ..... '
1
1
1
1
..1' 1 '■..... r
....
- ....
13C NMR Spectrum (100.0 MHz, D2 O solution)
D EP T
ch 2J[ ch3| cHf
proton decoupled
----------- 1----- 1----- 1----- 1------1----- 1----- 1----- 1------1----- 1------1------1------1----- j
200
10
378
9
160
8
7
120
6
5
80
4
1
1
1
40
3
2
1
1
0
1
1
8 (ppm)
0 8 (ppm)
Problem 268
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
i
i
|
i
r
T
T
T
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2iJ[ CH3f C H f
solvent
proton decoupled
j
1
i 1 200
i
1
i 1 160
i
i
i 120
.
i
J
80
1
L
40
0
8 (ppm)
5 (ppm)
379
Problem 269
100 7 80
28
1200
2000
1600 V (cm'1 )
3000
4000
Mass Spectrum
56
30
800
*
-. CO (D C 60 “- 0 0 l
.
CO _Q
40
7
O
20
7
No significant UV M+'
absorption above 220 nm
57
c 3 h 7n
Li! 40
80
120
160
200
240
280
m/ e I
1
'
1
'
'
1
'
1
1
1
I
1
'
1
1
I
i.....................................I i i i I i i. iI 80 40
i.
iI
I 0
I
1
I
I
1
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f CHf
solvent
proton decoupled
.............. ...................................................................................I i i i i I i i i I i
200
10
380
9
160
8
7
120
6
5
4
1
.
3
2
1
i______ 8(ppm)
0 5 (ppm)
.......
I-- ----'---1-- 1
I ' I ' I
Problem 270
IR Spectrum (CHCI3 solution) I
I
4000
.
1
.
«
.
I
.
I
3000
.
I----------------- |-------- 1-------- I--------.--------1--------1-------- 1--------1-------- 1-------- 1-------- 1-------
1200
2000 1600 V (crn1 )
L
800
100 80 No significant UV
60
absorption above 220 nm
40
20 40
80
120
160
200
240
280
m/e
I
!
I
i
I
.
I
I
I
I
I
I
I
I
I
I
| I
I
I
I
I
I
1----1----r
I
I
I
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH3f
c h
ch|
solvent proton decoupled
i
»
I
«
I
200
»
I
I
I
160
I
I
I
I
I
120
I
I
80
40
I
0
L
5 (ppm)
1H NMR Spectrum (400 MHz, CDCI 3 solution)
—
J1
A
i— --------r
1.9
1.7
ppm TMS
1
I. ---- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1 _
10
9
8
7
6
5
1
!
4
1
•
3
1
>
2
1
•
1
1
>
1
0 6 (ppm)
381
Problem 271
4000
3000
1200
2000 1600 V (crrf1 )
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e -i— i— 1— i— 1— r 13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
ch 2|
CH3f CHf
solvent
proton decoupled
I
10
382
■
I
■
l
8
■
I
I
I
I
I
I
I
I
.
I
I
I
I
I
I
I
I
I
I
1
L_
0 5 (ppm)
4000
3000
2000 1600 V (cm'1 )
1200
800
UV Spectrum
X max
281 nm (log10s 3.5)
X max 230 nm (log10s 3.6) solvent: methanol
40
80
120
160
200
240
280
m/ e 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
ch 2|
CH3f CHf
solvent
proton decoupled I
I
I
«
1
160
200
10
1
i _______ I_______ _______ I_______ I_______ I_______ I_______ I_______ I_______ I_______ L_
8
120
80
J
40
i
I
I
I
0
1
L_
5 (ppm)
0 8 (ppm)
383
Problem 273
100
Mass Spectrum
:
43
80 60
" .
CD q .
■ -
(D cf> CD
.
40
ro
-
No significant UV absorption above 220 nm
-Q
7 o - O" '"8
M+
20 :
’=
1 7 5 (< 1 % )
c 6h 9 n o 5
I i
*-----•----- *-----1----- •----- 1----- >-----1----- •-----1----- >----- •----- ■----- 1----- •----- 1----- •----- «----- •----- 1----- •----- 1----- ■----- 1----- •----- «----- •-----1----- 1___________ L
Chapter 10.3 Problems in 2D NMR
m/ e 1
1
1
' .-"T
1
'
1
'
A,(nm) 1
'
1
1
T"” 1
1
'
1
1 ”r
~r.....r .. •
1
1
13C NMR Spectrum (100 MHz, CDCI3 solution)
DEPT
ch 2{
CH3f CHf
solvent proton decoupled
1 — 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1____
200
160
120
80
40
0 5 (ppm)
6 (ppm)
454
Chapter 10.3 Problems in 2D NMR
1
JL
i
JL
--------- !----------1-------- 1-------- r*
JL
Ju
ppm
JUL
ppm
n----------- 1----------- 1----------- r
C-H Correlation Spectrum (1H 300 MHz; 13C 75 MHz; CDCI3 solution)
$
m
"i 4.0
i
i 3.0
r
-
20
-
40
-
60
n ------------------ 1-------------------1------------------ r
2.0
1.0
ppm
455
Chapter 10.3 Problems in 2D NMR
Problem 319 Use the basic spectral data plus the COSY and C-H correlation spectra on the facing page to deduce the structure of this compound.
4000
2000
3000
1200
1600
800
V (cm-1)
No significant UV absorption above 220 nm
40
120
80
160
200
240
280
m /e
200
5
160
4
120
3
80
40
2
0 8 (ppm)
1
0
8 (PPm)
456
Chapter 10.3 Problems In 2D NMR
I u k I ' 1 1 ' I 1 1 1' I 1 ' 11 I ' ' ' ' I 1 1 1 1 I
ppm
1H - 1H C O S Y Spectrum (500 MHz, CDCI3 solution)
1.0
2.0
4-
*
3.0
4.0
— I 1 1 1 1 I 1 1 1 1 I 1 11 1 I 1 11 1 I 1 1 11 I 1 1 1 1 I 4.0
3.0
2.0
J
1.0
u
J
ppm
l
I 1 " 1 I 1 11 1I
ppm
C-H Correlation Spectrum (1H 500 MHz; 13C 125 MHz; CDCI3 solution)
10
20
30
40
50
60
70
4.0
3.0
2.0
1.0
PPm
457
Chapter 10.3 Problems in 2D NMR
Problem 320 Use the basic spectral data and the NOESY spectrum on the facing page to deduce the structure of this compound.
X (nm)
200
160
80
120 solvent residual
1H NMR Spectrum
40
0 8 (ppm)
exchanges with D 20
*
(4 0 0 M H z, C 6 D 6 solution)
! »
expansion
7 .3
solvent residual
7.1
6 .9 ppm
6 .4
6 .2 ppm
exchanges with D 20
_J____ i
10
458
I___ i
9
L
8
J
L.
1
0 8 (ppm)
Chapter 10.3 Problems in 2D NMR
1H -1H NOESY Spectrum (300 MHz, Benzene-Dg solution) Diagonal peaks plotted with reduced intensity
ppm 3
3 ppm
459
Chapter 10.3 Problems in 2D NMR
Problem 321 Use the basic spectral data plus the COSY and NOESY spectra on the facing page to deduce the structure, including stereochemistry of this compound. 4000
3000
2000 1600 V (cm'1)
1200
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e I
1
I
1
I
1
I
1
I
1
I
1
1
1
'
1
'
1
'
1
'
1
1
1
'
1
I
1
I
1
I
1
1
13C NMR Spectrum (125 MHz, C D C I3 solution)
DEPT
CH 2)|( C H 3f C H f
solvent -ft proton decoupled
. 1 ______, i
1
I
200
1
I
1
I
160
i
I
1
1
120
1
I
1
I
80
1
I
40
0 §(ppm)
6 (ppm)
460
Chapter 10.3 Problems in 2D NMR
A_
-J
A_
Jj
lJ
461
Chapter 10.3 Problems in 2D NMR
Problem 322 Use the basic spectral data plus the COSY and NOESY spectra on the facing page to deduce the structure, including stereochemistry of this compound.
100:
Mass Spectrum 69
80
41
No significant UV
60
absorption above 220 nm
40 M+‘
20
136
jiuL
154
Ci 0 Hi8O
I. _j
40
80
120
160
i
i
i
200
i
■
i
.
240
i
i
i_
280
m/ e
13C NMR Spectrum (125 MHz, CDCI3 solution)
DEPT
CH3t
c h 2|
c h
f
solvent proton decoupled
J
.
I
200
L
_ l
.
I
160
L_
J
120
1
I
I
I
80
I
I
I
I
40
1
I
L-
0 8 (ppm)
5 (ppm)
462
Chapter 10.3 Problems in 2D NMR
A
A
___
A
A
___
ppm
2.0
3.0
4.0
5.0
6.0
463
Chapter 10.3 Problems in 2D NMR
Problem 323 U se the basic spectral data and the C-H Correlation and HMBC spectra on the the following pages to determine the structure of the compound, including its stereochemistry. 4000
100 80 60
■ • . “■ -
3000
2000 1600 V (cm'1)
800
Mass Spectrum
203/205
UV Spectrum log10s >4.5
CD - solvent
r~i- solvent ......J..... proton decoupled
130 _______ __ ___________ I...........
160
i
120
1H NMR Spectrum
-
ppm
I I
1
— i----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1— i----- 1
200
1 2 0
J.
l
i
i
i
80
i
i
i
i
i
40
i
i
0
§(ppm)
expansion
(400 MHz, Acetone-d6 solution)
J U l
7.4
6.2 ppm
6.6
7.0 expansion scale 0
40 80 120 Hz
solvent residual TMS J _____
10
464
9
8
l
L_
7
6
5
4
3
2
. 1
I
i 0 5 (ppm)
Chapter 10.3 Problems in 2D NMR
C-H Correlation Spectrum solvent residual
( 1H 400 MHz; 13C 100 MHz; Acetone-d6 solution)
JL
u___________________
u
ppm
solvent
40
60
- 80
100
120 -*• -l-i-
- 140
II
ppm
II — rn-------- 1--------- 1--------- 1-------- 1— C-H Correlation Spectrum (expansion) ( 1H 400 MHz; 13C 100 MHz; Acetone-d6 solution)
ppm
00 120
130
00
140
- 150
I 7.4
7.0
6.6
ppm
465
Chapter 10.3 Problems in 2D NMR
C-H HMBC Spectrum ( 1H 400 MHz; 13C 100 MHz; Acetone-d6 solution)
Jl
solvent
466
II
II____________________________________
solvent residual
10.4 NMR SPECTRAL ANALYSIS
46”
Chapter 10.4 NMR Spectral Analysis
Problem 324 Give the number of different chemical environments for the magnetic nuclei 'H and l3C in the following compounds. Assume that any conformational processes are fast on the NMR timescale unless otherwise indicated. Structure CH3 -C O -C H 2CH2CH3 CH 3CH 2 - CO - CH 2CH 3 c h 2= c h c h 2c h 3
cis- CH3CH=CHCH3 frans-CH3CH=CHCH3
0
0
O O ci a: Br—
^ — Br
Br —
Cl
C l^ y O C H s
Assuming slow chairchair interconversion Assuming fast chairchair interconversion H
Assuming the molecule to be
conformationally rigid
468
Number of 1H environments
Number of 13C environments
Chapter 10.4 NMR Spectral Analysis
Problem 325 Draw a schematic (line) representation of the pure first-order spectrum (AMX) corresponding to the following parameters: Frequencies (Hz from TMS):
vA= 300; vM= 240; vx =120.
Coupling constants (Hz):
JAM = 10; JAX = 2; JMX = 8 .
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 6 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
_l
350
300
250
I I I \
-----,-----1-----(-----1-
200
150
100
(Hz from TMS)
469
Chapter 10.4 NMR Spectral Analysis
Problem 326 Draw a schematic (line) representation of the pure first-order spectrum (AMX) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 180; vM= 220; vx = 300.
Coupling constants (Hz):
Jam = 10; JAX = 12 ; J MX = 5.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 8 scale corresponding to the above spectrum obtained with an instrument operating at 200 MHz for protons.
-- 1--- j--- ,--- 1--- 1 ----1--- 1 --- 1----1--- 1--- 1--- 1--- 1----1 --- 1--- 1--- ,--- 1----1--- 1 --- 1--- 1--- 1----1--- 1--- 1 --- 1--- 1----1—
350
470
300
250
200
150
100
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 327 Draw a schematic (line) representation of the pure first-order spectrum (AX2) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 150; vx = 300.
Coupling constants (Hz):
*^AX 2 0 .
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 8 scale corresponding to the above spectrum obtained with an instrument operating at 400 MHz for protons.
-I 1 r 350
-1--- 1--- r 300
250
200
\ 1 1 r 150
_|
100
,
,
f_
(Hz from TMS)
471
Chapter 10.4 NMR Spectral Analysis
Problem 328 A 60 MHz 'H NMR spectrum of diethyl ether is given below. Note that the spectrum is calibrated only in parts per million (ppm) from tetramethylsilane (TMS), i.e. in 5 units. (a)
Assign the signals due to the -C H 2- and -C H 3 groups respectively using three independent criteria (the relative areas of the signals, the multiplicity of each signal and the chemical shift of each signal).
(b)
Obtain the chemical shift of each group in ppm, then convert to Hz at 60 MHz from TMS (see Section 5.5).
(c)
Obtain the value of the first-order coupling constants 3J H.H (in Hz).
(d)
Demonstrate that first-order analysis was justified (see Section 5.9).
TMS
Jm
J
t 3.0
472
2.0
1.0
0.0 8 ppm from TMS
Chapter 10.4 NMR Spectral Analysis
Problem 329 A 100 MHz 'H NMR spectrum of a 3-proton system is given below. (a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (5 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
1H 500
1H 450
1H 400 (Hz from TMS)
473
Chapter 10.4 NMR Spectral Analysis
Problem 330 Portion of the 60 MHz NMR spectrum 2-furoic acid in CDCI3 is shown below. Only the resonances due to the three aromatic protons (Ha, Hm and Hx) are shown.
/ Ha "
\ O
2-Furoic Acid COOH
(a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (8 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
Note: This is a 60 MHz spectrum.
H„
474
H„
Chapter 10.4 NMR Spectral Analysis
Problem 331 COOH
A portion of the 100 MHz 'H NMR spectrum of 2-amino-5-chlorobenzoic acid in CD3OD is given below. Only the resonances due to the three aromatic protons are shown.
H4
(a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (5 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
(c)
Assign the three multiplets to H3, H4 and H 6 given: •
the characteristic ranges for coupling constants between aromatic protons (see Section 5.9);
•
the fact that H3 will give rise to the resonance at the highest field due to the strong influence of the amino group (see Table 5.6).
475
Chapter 10.4 NMR Spectral Analysis
Problem 332a Portion of 100 MHz 'H NMR spectrum of methyl acrylate (5% in CeDg) is given below. Only the part of the spectrum containing the resonances of the olefinic protons Ha, H b and Hc is shown.
/ , . / C = C V. Hb He methyl acrylate
476
(a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (5 in ppm) by direct measurement.
(c)
Justify the statement that "this spectrum is really a borderline second-order (strongly coupled) case". Point out the most conspicuous deviation from first-order character in this spectrum (see Section 5.9).
(d)
Assign the three multiplets to HA, HB and Hc on the basis of coupling constants only (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 332b This is the computer-simulated spectrum corresponding to the complete analysis of the spectrum shown in Problem 318a, i.e. an exact analysis in which first-order assumptions were not made. The simulated spectrum fits the experimental spectrum, verifying that the analysis was correct. Compare your (first-order) results from Problem 3 18a with the actual solution given here. Number of SPINS F(l) F(2) F(3) J(l,2) J(l,3) J(2,3)
3 = + 528.500 Hz = + 594.531 Hz = + 626.093 Hz + 10.539 Hz + 1.589 Hz + 17.278 Hz
START of simulation = + 750.000 Hz FINISH of simulation = + 500.000 Hz LINE WIDTH = + 0.427 Hz
Chapter 10.4 NMR Spectral Analysis
Problem 333 Draw a schematic (line) representation of the pure first-order spectrum (AX3) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 160; vx = 280.
Coupling constants (Hz):
JAX =15.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 5 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
I
350
478
I
I
I
I
I
300
I
I
I
I
I
250
I
I
I
I
I
200
I
I
I
I
I
150
I
I
I
I
I
100
I
I
I
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 334 A 100 MHz *H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
U l 3H 200
1H 150
100
(Hz from TMS)
479
Chapter 10.4 NMR Spectral Analysis
Problem 335 Draw a schematic (line) representation of the pure first-order spectrum (AMX2) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 340; vM= 240; vx = 100.
Coupling constants (Hz):
J AM = 10; JAX = 2; J MX = 6 .
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 5 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
i
350
480
1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-
300
250
200
150
100
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 336 Draw a schematic (line) representation of the pure first-order spectrum (AM2X) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 110; vM= 200; vx = 290.
Coupling constants (Hz):
JAM = 10; J AX = 12; J MX = 3.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 8 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
I
350
I
I
I
“I
I
300
I
I
I
I
I
250
I
I
I
1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1—
200
150
100
(Hz from TMS)
481
Chapter 10.4 NMR Spectral Analysis
Problem 337 A 100 MHz 'H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (5 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
I 2H
1H 600
482
500
1H 400
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 338 A 100 MHz 'H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 339 A 100 MHz 'H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 340 A 200 MHz 'H NMR spectrum of a 5-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
2H
2H 1010
1H
— r~
—i—
990
Hz from TMS
510
490 Hz from TMS
90
110 Hz from TMS
485
Chapter 10.4 NMR Spectral Analysis
Problem 341 COOH
Portion of 100 MHz NMR spectrum of crotonic acid in CDCI3 is given below. The upfield part of the spectrum, which is due to the methyl group, is less amplified to fit the page.
Ch/
c=c \ Hm crotonic acid
(a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (5 in ppm) by direct measurement. Justify the use of first-order analysis.
(b)
There are certain conventions used for naming spin-systems {e.g. A M X , AMX2, AM2X3). Note that this is a 5-spin system and name the spin system responsible for this spectrum (see Section 5.9).
730
720
710
700
690
600
590
580
(Hz from TMS)
486
570
200
190
180
Chapter 10.4 NMR Spectral Analysis
Problem 342 The 100 MHz 'H NMR spectrum (5% in CDC13 ) of an a,(3-unsaturated aldehyde C4H60 is given below. (a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
(c)
Use the coupling constants to obtain the structure of the compound, including the stereochemistry about the double bond (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 343 Draw a schematic (line) representation of the pure first-order spectrum (AMX3) corresponding to the following parameters: Frequencies (Hz from TMS):
vA= 80; vM= 220; vx = 320.
Coupling constants (Hz):
JAM = 10; J AX = 12; J MX = 0.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 5 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
1
350
488
1
1
1
1
1
300
1
1
1
1
1
250
1
1
1
1
1
200
1
1
1
1
1
150
1
1
1
1
1
100
1
1
1-
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 344 A portion of the 90 MHz *H NMR spectrum (5% in CDCI3) of one of the six possible isomeric dibromoanilines is given below. Only the resonances of the aromatic protons are shown. Br
Br Br
Br
NH2
NHo
nh2
NH2
Br
Br Br
Br
Determine which is the correct structure for this compound using arguments based on symmetry and the magnitudes of spin-spin coupling constants (see Section 5.9).
4^9
Chapter 10.4 NMR Spectral Analysis
Problem 345 The 400 MHz 'H NMR spectrum (5% in CDCI3 after D2O exchange) of one of the six possible isomeric hydroxycinnamic acids is given below. OH
OH
OH
Determine which is the correct structure for this compound using arguments based on symmetry and the magnitudes of spin-spin coupling constants (see Section 5.9).
1H
490
Chapter 10.4 NMR Spectral Analysis
Problem 346 In a published paper, the 90 MHz 'H NMR spectrum given below was assigned to 1,5-dichloronaphthalene, C,„H6C12. 1,5-dichloronaphthalene
(a)
Why can't this spectrum belong to 1,5-dichloronaphthalene?
(b)
Suggest two alternative dichloronaphthalenes that would have structures consistent with the spectrum given.
491
Chapter 10.4 NMR Spectral Analysis
Subject Index
Subject Index Key:
13C NM R = Carbon 13 nuclear magnetic resonance spectroscopy NM R = Proton nuclear magnetic resonance spectroscopy 2D NM R = 2-dimensional NMR IR = Infrared spectroscopy MS = Mass spectrometry UV = Ultraviolet spectroscopy
Absorbance, molar
8, 9
Aldehydes
Anisotropy, magnetic, NM R
50,51
Appearance potential, MS
21
13C NMR
76
'H NM R
45, 46
,3C NM R
76, 80
IR
17,18
coupling constants, NMR
68
'H N M R
4 5 ,4 9 ,5 0
Aromatic compounds
Alkanes l3C NM R
76, 78
polynuclear
4 9 ,80
'H NMR
45, 46, 66
UV
13
Alkenes
Aromatic Solvent Induced Shift
94
l3C N M R
76,78
(ASIS)
'H N M R
4 5 ,4 8 ,6 7
Auxochrome,
10, 13
IR
19,20
Base peak, MS
24
UV
11
Bathochromic shift, UV
10
Beer-Lambert Law
2, 8
Alkynes l3C NMR
76, 79
Boltzmann excess, NM R
36
'H N M R
45 ,4 8
Cation radical, MS
21,22
IR
19,20
Carbonyl compounds
Allenes
i3C N M R
76-80
l3C N M R
75
IR
17,18
IR
20
MS
32,33
UV
12
Amides l3C NMR
76
'H N M R
4 6 ,5 1 ,9 8
l3C NM R
76, 77
IR
18
'H N M R
45,51
IR
18
MS
33
Amines 'H N M R
4 6 ,5 1 ,9 7
IR
19
Analysis o f 'H NMR Spectra Anion Radical, MS
Carboxylic acids
Chemical Ionisation, MS
22
56-65 21
Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
493
Subject Index
Chemical shift aromatic solvent induced
42 94
(ASIS)
i3C N M R
76
IR
18
MS
32,33
l3C, tables
75-80
factors influencing
42-44,
Exchange broadening, NM R
96
50, 51
Exchangeable protons, NM R
51, 52, 97
]H, tables
45-49
l9F N M R
35,99
scale
43 ,44
First-order spectra, NMR
57
standard
42,43
rules for analysis
Chirality, effect on NM R
98,99
Fourier transformation, NM R
40
Chromophore
3
Fourier Transform Infrared, FTIR
16
Fragmentation, MS
21,26-33
Cleavage, MS
56, 59-60
a-
32
common fragments
27
p-
31
Free induction decay (FID), NMR
40
Conformational exchange
97,98
Halogen derivatives, IR
19
Connectivity
4
Halogen derivatives, isotopes, MS
27-29
Contour plot, 2D NMR
82
Heteroaromatic compounds
Correlation Spectroscopy (COSY)
83,84
2D NM R
i3C N M R
80
'HNMR
49
Heteronuclear Single Quantum
Coupling constant
494
Esters,
82, 84-85
NMR
53
Correlation (HSQC), 2D NMR
allylic
67
Heteronuclear Shift Correlation
aromatic systems
68-70
(HSC), 2D NMR
geminal
66
High-resolution mass spectroscopy
24-25
heteroaromatic systems
68
Hydrogen bonding, IR
17
long range
67,88
Hydroxyl groups
olefmic
67
IR
17
vicinal
66
'H N M R
4 5 ,5 1 ,9 7
Cyanates, IR
20
Hypsochromic shift, UV
10
Degree o f Unsaturation
3 ,4
Imines, IR
19
Deshielding, NMR
4 4 ,5 0 ,5 1
Intermolecular exchange
97
Dienes, UV
11
Ionisation, MS
D20 exchange
51,52
chemical ionisation (Cl)
22
DEPT, l3C NMR
72,73
electron impact, (El)
21-23
Electrospray Ionisation (ESI), MS
22
electrospray (ESI)
22
Enol ethers, IR
19
matrix assisted (MALDI)
22
Equivalence, NMR
4 3 ,57
Isotope ratio, MS
27-29
accidental
43
Isocyanates, IR
20
chemical
4 3 ,57
Karplus relationship, NM R
66
magnetic
57
82, 84-85
Subject Index
Ketones
Saturation, NM R
37
13C NMR
76
Sensitivity
5
IR
18
Shielding, NM R
4 3 ,5 0 ,5 1
MS
32,33
Solvents for NM R
95
4 5 ,5 1 ,9 7
Spectrometry, Mass
21
Labile protons
Spectroscopy, definition of
Lactones
1
IR
18
IR
15
Larmor equation
35
NM R
34
M + 1, M+2 peaks, MS
27-28
,3C N M R
71
Magnetic anisotropy
50-51
continuous wave (CW)
39
M cLafferty rearrangement
33
Fourier transform (FT)
40
MALDI, MS
22
UV
7
Mass number, MS
24
pH dependence
13-14
Mass spectrometry
21
solvent dependence
14
Matrix Assisted Laser Desorption
22
Ionisation, (MALDI) MS
Spin, nuclear, NM R
34
Spin decoupling, NMR
65
Metastable peaks, MS
29
broadband
71
Molecular ion, MS
21
noise
71
Nitrogen Rule, MS
24
selective
65
Nitriles
Spin quantum number, NMR
34, 35
13C NMR
76-78,80
'H N M R
46-49
strongly coupled systems
56,57
IR
20
weakly coupled systems
56,57
Spin-spin coupling
53
Nitro compounds, IR
19
NM R spectroscopy
34,71
NM R time-scale
96-98
Splitting diagram, NM R
61
Nuclear Overhauser effect (NOE)
41
Structural element
3
Sulfonamides, IR
19
NM R
Spin system, NM R naming conventions
56 57
NOESY, 2D NMR
82,91-92
Sulfonate esters, IR
19
31p n m r
35,99
Sulfones, IR
19
Partial double bonds
98
Sulfoxides, IR
19
Time o f Flight (TOF), MS
24
Polynuclear aromatic compounds 13c n m r
80
Two-dimensional NMR
81
'h n m r
4 6 ,4 9 ,6 8
T 1? NMR
37
Prochiral centre
98-99
Thiocyanates, IR
20
Relaxation, NM R
37
Thiols, 'H N M R
4 5 ,5 1 ,9 7
spin lattice
37
TOCSY, 2D NM R
82,92-93
Residual solvent peaks
94-95
Wavenumber, IR
15
Resonance, NM R
35
Ring current effect, NM R
50
495
i
ii
Organic Structures from Spectra Fifth Edition L. D. Field University o f New South Wales, Australia
S. Sternhell University o f Sydney, Australia
J. R. Kalman University o f Technology Sydney, Australia
iii
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iv
CONTENTS PREFACE LIST OF TABLES LIST OF FIGURES 1 INTRODUCTION 1.1 1.2 1.3 1.4 1.5 1.6
2
ULTRAVIOLET (UV) SPECTROSCOPY 2.1 2.2 2.3 2.4 2.5 2.6 2.7
3
ABSORPTION RANGE AND THE NATURE OF IR ABSORPTION EXPERIMENTAL ASPECTS OF INFRARED SPECTROSCOPY GENERAL FEATURES OF INFRARED SPECTRA IMPORTANT IR CHROMOPHORES
MASS SPECTROMETRY 4.1 4.2 4.3 4.4 4.5 4.6
5
BASIC INSTRUMENTATION THE NATURE OF ULTRAVIOLET SPECTROSCOPY QUANTITATIVE ASPECTS OF ULTRAVIOLET SPECTROSCOPY CLASSIFICATION OF UV ABSORPTION BANDS SPECIAL TERMS IN ULTRAVIOLET SPECTROSCOPY IMPORTANT UV CHROMOPHORES THE EFFECT OF SOLVENTS
INFRARED (IR) SPECTROSCOPY 3.1 3.2 3.3 3.4
4
GENERAL PRINCIPLES OF ABSORPTION SPECTROSCOPY CHROMOPHORES DEGREE OF UNSATURATION CONNECTIVITY SENSITIVITY PRACTICAL CONSIDERATIONS
IONISATION PROCESSES INSTRUMENTATION MASS SPECTRAL DATA REPRESENTATION OF FRAGMENTATION PROCESSES FACTORS GOVERNING FRAGMENTATION PROCESSES EXAMPLES OF COMMON TYPES OF FRAGMENTATION
NUCLEAR MAGNETIC RESONANCE(NMR) SPECTROSCOPY 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9 5.10
THE PHYSICS OF NUCLEAR SPINS AND NMR INSTRUMENTS CONTINUOUS WAVE (CW) NMR SPECTROSCOPY FOURIER-TRANSFORM (FT) NMR SPECTROSCOPY THE NUCLEAR OVERHAUSER EFFECT (NOE) CHEMICAL SHIFT IN 1H NMR SPECTROSCOPY SPIN-SPIN COUPLING IN 1H NMR SPECTROSCOPY ANALYSIS OF 1H NMR SPECTRA CHANGING THE MAGNETIC FIELD IN NMR SPECTROSCPY RULES FOR SPECTRAL ANALYSIS OF FIRST ORDER SPECTRA CORRELATION OF 1H - 1H COUPLING CONSTANTS WITH STRUCTURE
vii xi xiii 1 1 3 3 4 5 5 7 7 8 8 9 10 10 14 15 15 16 16 17
21 21 23 24 29 30 30 34 34 39 40 41 42 53 56 58 59 66
V
Contents
6 13C NMR SPECTROSCOPY 6.1 6.2 6.3
COUPLING AND DECOUPLING IN 13C NMR SPECTRA DETERMINING 13C SIGNAL MULTIPLICITY USING DEPT SHIELDING AND CHARACTERISTIC CHEMICAL SHIFTS IN 13C NMR SPECTRA
7 2-DIMENSIONAL NMR SPECTROSCOPY 7.1 7.2
7.3 7.4 7.5
COSY (CORRELATION SPECTROSCOPY) THE HSQC (HETERONUCLEAR SINGLE QUANTUM CORRELATION) OR HSC (HETERONUCLEAR SHIFT CORRELATION) SPECTRUM HMBC (HETERONUCLEAR MULTIPLE BOND CORRELATION) NOESY (NUCLEAR OVERHAUSER EFFECT SPECTROSCOPY) TOCSY (TOTAL CORRELATION SPECTROSCOPY)
8 MISCELLANEOUS TOPICS 8.1 8.2 8.3 8.4 8.5
SOLVENTS FOR NMR SPECTROSCOPY SOLVENT INDUCED SHIFTS DYNAMIC PROCESSES IN NMR ■THE NMR TIME-SCALE THE EFFECT OF CHIRALITY THE NMR SPECTRA OF "OTHER NUCLEI"
9 DETERMINING THE STRUCTURE OF ORGANIC COMPOUNDS FROM SPECTRA 9.1 9.2
71 72 75
81 83 84
86 91 92
94 94 95 96 98 99
100 102 103
10 PROBLEMS
111
10.1 10.2 10.3 10.4
111 397 407 467
INDEX
VI
SOLVING PROBLEMS WORKED EXAMPLES
71
SPECTROSCOPIC IDENTIFICATION OF ORGANIC COMPOUNDS THE ANALYSIS OF MIXTURES PROBLEMS IN 2-DIMENSIONAL NMR NMR SPECTRAL ANALYSIS
493
PREFACE
The derivation of structural information from spectroscopic data is an integral part of Organic Chemistry courses at all Universities. At the undergraduate level, the principal aim of courses in organic spectroscopy is to teach students to solve simple structural problems efficiently by using combinations of the major techniques (UV, IR, NMR and MS). Over a period more than 30 years, we have evolved courses at the University of Sydney and at the University of New South Wales, which achieve this aim quickly and painlessly. The text is tailored specifically to the needs and philosophy of these courses. As we believe our approach to be successful, we hope that it may be of use in other institutions. The courses has been taught at the beginning of the third year, at which stage students have completed an elementary course of Organic Chemistry in first year and a mechanistically-oriented intermediate course in second year. Students have also been exposed, in their Physical Chemistry courses, to elementary spectroscopic theory, but are, in general, unable to relate the theory to actually solving spectroscopic problems. We have delivered courses of about 9 lectures outlining the basic theory, instrumentation and the structure-spectra correlations of the major spectroscopic techniques. The text of this book broadly corresponds to the material presented in the 9 lectures. The treatment is both elementary and condensed and, not surprisingly, the students have great difficulties in solving even the simplest problems at this stage. The lectures are followed by a series of 2-hour problem solving seminars with 5 to 6 problems being presented per seminar. At the conclusion of the course, the great majority of the class is quite proficient and has achieved a satisfactory level of understanding of all methods used. Clearly, the real teaching is done during the hands-on problem seminars, which are organised in a manner modelled on that which we first encountered at the E.T.H. Zurich. The class (typically 60 - 100 students, attendance is compulsory) is seated in a large lecture theatre in alternate rows and the problems for the day are identified. The students are permitted to work either individually or in groups and may use any written or printed aids they desire. Students solve the problems on their individual copies of this book thereby transforming it into a set of worked examples and most students voluntarily complete many more problems than are set. Staff (generally 4 or 5) wander around giving help and tuition as needed - the empty alternate rows of
vii
Preface
seats make it possible to speak to each student individually. When an important general point needs to be made, the staff member in charge gives a very brief exposition at the board. There is a 1½ hour examination consisting essentially of 4 problems from the book and the results are in general very satisfactory. Moreover, the students themselves find this a rewarding course since the practical skills acquired are obvious to them. Solving these real puzzles is also addictive - there is a real sense of achievement, understanding and satisfaction, since the challenge in solving the graded problems builds confidence even though the more difficult examples are quite demanding. Our philosophy can be summarised as follows: (a)
Theoretical exposition must be kept to a minimum, consistent with gaining of an understanding of the parts of the technique actually used in solving the problems. Our experience indicates that both mathematical detail and description of advanced techniques merely confuse the average student.
(b)
The learning of data must be kept to a minimum. We believe that it is more important to learn to use a restricted range of data well rather than to achieve a nodding acquaintance with more extensive sets of data.
(c)
Emphasis is placed on the concept of identifying "structural elements" and the logic needed to produce a structure out of the structural elements.
We have concluded that the best way to learn how to obtain "structures from spectra" is to practise on simple problems. This book was produced principally to assemble a suitable collection of problems for that purpose. Problems 1-282 are of the standard “structures from spectra” type and are arranged roughly in order of increasing difficulty. A number of problems deal with related compounds (sets of isomers) which differ mainly in symmetry or the connectivity of the structural elements and are ideally set together. The sets of related examples include: problems 3 and 4; 19 and 20; 31 and 32; 42 and 43; 44, 45 and 46; 47, 48 and 49; 50 and 51; 61, 62 and 63; 64, 65 and 6 6 ; 81 and 82; 84 and 85; 99, 100, 101 and 102; 107 and 108; 1 10, 111, 112 and 113; 114 and 115; 118, 119 and 120; 122 and 123; 127 and 128; 139, 140, 141, 142 and 143; 155, 156, 157, 158, 159 and 160; 179 and 180; 181 and 182; 185 and 186; 215 and 216; 226 and 227; 235, 236 and 237; 276 and 277. A further group of problems offer practice in the analysis of proton NMR spectra: 19, 20, 29, 37, 58, 75, 79, 90, 92, 93, 94, 99, 101, 123, 137, 146, 159, 163, 164, 183, 187, 192, 195, 205, 208, 236, 237, 238, 239, 248, 250, 251, 252 and 260.
viii
Preface
A number of problems (195, 196, 197, 198, 230, 231, 260, 264, 265, 268, 271, 274 and 275) exemplify complexities arising from the presence of chiral centres, or from restricted rotation about peptide bonds (128, 162 and 262), while some problems deal with structures of compounds of biological, environmental, or industrial significance (22, 23, 36, 8 6 , 95, 127, 131, 132, 144, 153, 162, 164, 197, 204, 220, 259, 260, 261, 263, 264, 265, 267, 272, 273, 274 and 275). Problems 283-288 are again structures from spectra, but with the data presented in a textual form such as might be encountered when reading the experimental section of a paper or report. Problems 289-296 deal with the use of NMR spectroscopy for quantitative analysis and for the analysis of mixtures of compounds. Problems 297-323 represent a considerably expanded set of problems dealing with the interpretation of two-dimensional NMR spectra and are a series of graded exercises utilising COSY, NOESY, C-H Correlation, HMBC and TOCSY spectroscopy as aids to spectral analysis and as tools for identifying organic structures from spectra. Problems 324-346 deal specifically with more detailed analysis of NMR spectra, which tends to be a stumbling block for many students. In Chapter 9, there are also two worked solutions (to problems 96 and 127) as an illustration of a logical approach to solving problems. However, with the exception that we insist that students perform all routine measurements first, we do not recommend a mechanical attitude to problem solving - intuition has an important place in solving structures from spectra as it has elsewhere in chemistry.
Bona fide instructors may obtain a list of solutions (at no charge) by writing to the authors or EMAIL: [email protected] or FAX: (61-2)-9385-8008
We wish to thank Dr Alison Magill, and Dr Hsiu Lin Li in the School of Chemistry at the University of New South Wales and Dr Ian Luck at the University of Sydney who helped to assemble the many additional samples and spectra in the 4th and 5th editions of this book. Thanks are also due to the many graduate students and research associates who, over the years, have supplied us with many of the compounds used in the problems. L. D. Field S. Sternhell J. R. Kalman September 2012
ix
X
LIST OF TABLES Table 2.1
The Effect of Extended Conjugation on UV Absorption
11
Table 2.2
UV Absorption Bands in Common Carbonyl Compounds
12
Table 2.3
UV Absorption Bands in Common Benzene Derivatives
13
Table 3.1
Carbonyl (C =0) IR Absorption Frequencies in Common Functional Groups
18
Table 3.2
Characteristic IR Absorption Frequencies for Functional Groups
19
Table 3.3
Common IR Absorption Frequencies in the Region 1900 - 2600 cm '1
20
Table 4.1
Accurate Masses of Selected Isotopes
25
Table 4.2
Common Fragments and their Masses
27
Table 5.1
Resonance Frequencies of ]H and 13C Nuclei in Magnetic Fields of Different Strengths
36
Table 5.2
Typical
45
Table 5.3
Typical lR Chemical Shift Ranges in Organic Compounds
Table 5.4
Chemical Shift Values in Selected Organic Compounds
Chemical Shifts (8) for Protons in Common Alkyl Derivatives
45 46
Table 5.5
Approximate lH Chemical Shifts (8) for Olefinic Protons C=C-H in ppm
48
Table 5.6
Approximate lH Chemical Shifts (8) for Aromatic Protons in Benzene Derivatives Ph-X in ppm Relative to Benzene at 8 7.26 ppm
49
Table 5.7
]H Chemical Shifts (8) in some Polynuclear Aromatic Compounds and Heteroaromatic Compounds
49
Table 5.8
Typical lR -
54
Table 5.9
Relative Line Intensities for Simple Multiplets
54
Table 6.1
The Number of Aromatic 13C Resonances in Benzenes with Different Substitution Patterns
74
Table 6.2
Typical 13C Chemical Shift Values in Selected Organic Compounds
75
Table 6.3
Typical 13C Chemical Shift Ranges in Organic Compounds
76
Table 6.4
13C Chemical Shifts (8) for sp3 Carbons in Alkyl Derivatives
78
Table 6.5
13C Chemical Shifts (8) for sp2 Carbons in Vinyl Derivatives CH2=CH-X
78
Table 6.6
13C Chemical Shifts (8) for sp Carbons in Alkynes: X-C=C-Y
79
Table 6.7
Approximate 13C Chemical Shifts (8) for Aromatic Carbons in Benzene Derivatives Ph-X in ppm relative to Benzene at 8 128.5 ppm
80
Table 6.8
Characteristic 13C Chemical Shifts (8) in some Polynuclear Aromatic Compounds and Heteroaromatic Compounds
80
Table 8.1
]H and 13C Chemical Shifts for Common NM R Solvents
95
Coupling Constants
xi
xii
LIST OF FIGURES Figure 1.1
Schematic Absorption Spectrum
1
Figure 1.2
Definition o f a Spectroscopic Transition
2
Figure 2.1
Schematic Representation o f an IR or UV Spectrometer
7
Figure 2.2
Definition of Absorbance (A)
9
Figure 4.1
Schematic Diagram of an Electron-Impact Mass Spectrometer
23
Figure 4.2
Relative Intensities of the Cluster of Molecular Ions for Molecules Containing Combinations of Bromine and Chlorine Atoms
29
Figure 5.1
A Spinning Charge Generates a Magnetic Field and Behaves Like a Small Magnet
34
Figure 5.2
Schematic Representation of a CW NM R Spectrometer
39
Figure 5.3
Time Domain and Frequency Domain NM R Spectra
40
Figure 5.4
Approximate ]H Chemical Shift Ranges for Protons in Organic Compounds
47
Figure 5.5
Shielding/deshielding Zones for Common Non-aromatic Functional Groups
51
Figure 5.6
D20 Exchange in the *H Spectrum of 1-Propanol
52
Figure 5.7
Characteristic Multiplet Patterns for Common Organic Fragments
55
Figure 5.8
A Portion o f the solution in CC14)
61
Figure 5.9
The 60 MHz *H NMR Spectrum of a 4-Spin AMX2 Spin System
NMR Spectrum o f Styrene Epoxide (100 MHz as a 5%
62
Figure 5.10 Simulated lH NMR Spectra of a 2-Spin System as the Ratio Av/J, is Varied from 10.0 to 0.0
63
Figure 5.11
Selective Decoupling in a Simple 4-Spin System
65
Figure 5.12
Characteristic Aromatic Splitting Patterns in the !H NM R spectra for some Diand Tri-substituted Benzenes
69
Figure 5.13
jH NM R Spectrum ofj^-Nitrophenylacetylene (200 MHz as a 10% solution in CDC13)
70
Figure 6.1
13C NM R Spectra of Methyl Cyclopropyl Ketone (CDC13 Solvent, 100 MHz). (a) with Broad Band Decoupling of *H; (b) DEPT Spectrum (c) with no Decoupling of *H
73
Figure 6.2
Approximate 13C Chemical Shift Ranges for Carbon Atoms in Organic Compounds
77
Figure 7.1
]H COSY Spectrum of 1-Iodobutane
83
Figure 7.2
]H - 13C HSQC Spectrum of 1-Iodobutane
85
Figure 7.3
]H - 13C HMBC Spectrum of 1-Iodobutane
87
Figure 7.4
*H - 13C HMBC Spectrum of 2-Bromophenol
89
Figure 7.5
]H NOESY Spectrum of (3-Butyrolactone
91
Figure 7.6
]H TOCSY Spectrum of Butyl Ethyl Ether
93
Figure 8.1
Schematic NMR Spectra of Two Exchanging Nuclei
96
Figure 8.2
]H NM R Spectrum of the Aliphatic Region o f Cysteine
99
xiv
1 INTRODUCTION
1.1
GENERAL PRINCIPLES OF ABSORPTION SPECTROSCOPY
The basic principles of absorption spectroscopy are summarised below. These are most obviously applicable to UV and IR spectroscopy and are simply extended to cover NMR spectroscopy. Mass Spectrometry is somewhat different and is not a type of absorption spectroscopy. Spectroscopy is the study of the quantised interaction of energy (typically electromagnetic energy) with matter. In Organic Chemistry, we typically deal with molecular spectroscopy i.e. the spectroscopy of atoms that are bound together in molecules. A schematic absorption spectrum is given in Figure 1.1. The absorption spectrum is a plot of absorption of energy (radiation) against its wavelength (X) or frequency (v).
AE
Figure 1.1
Schematic Absorption Spectrum
Organic Structures from Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
Chapter 1 Introduction
An absorption band can be characterised primarily by two parameters: (a)
the wavelength at which maximum absorption occurs
(b)
the intensity of absorption at this wavelength compared to base-line (or background) absorption
A spectroscopic transition takes a molecule from one state to a state of a higher energy. For any spectroscopic transition between energy states (e.g. E, and E2 in Figure 1.2), the change in energy (AE) is given by: AE = hv where h is the Planck's constant and v is the frequency of the electromagnetic energy absorbed. Therefore v oc AE.
A
----------------
I
t AE = E2 - Ei
Energy
-------Figure 1.2
e2
Ei
Definition of a Spectroscopic Transition
It follows that the x-axis in Figure 1.1 is an energy scale, since the frequency, wavelength and energy of electromagnetic radiation are interrelated: vX = c (speed of light) X=±-
v
iA 1 , oc----
AE A spectrum consists of distinct bands or transitions because the absorption (or emission) of energy is quantised. The energy gap of a transition is a molecular property and is characteristic o f molecular structure. The y-axis in Figure 1.1 measures the intensity of the absorption band and this depends on the number of molecules observed (the Beer-Lambert Law) and the probability of the transition between the energy levels. The absorption intensity is also a molecular property and both the frequency and the intensity of a transition can provide structural information.
2
Chapter 1 Introduction
1.2
CHROMOPHORES
In general, any spectral feature, i.e. a band or group of bands, is due not to the whole molecule, but to an identifiable part of the molecule, which we loosely call a chromophore. A chromophore may correspond to a functional group (e.g. a hydroxyl group or the double bond in a carbonyl group). However, it may equally well correspond to a single atom within a molecule or to a group of atoms (e.g. a methyl group) which is not normally associated with chemical functionality. The detection of a chromophore permits us to deduce the presence of a structural fragment or a structural element in the molecule. The fact that it is the chromophores and not the molecules as a whole that give rise to spectral features is fortunate, otherwise spectroscopy would only permit us to identify known compounds by direct comparison of their spectra with authentic samples. This "fingerprint" technique is often useful for establishing the identity of known compounds, but the direct determination of molecular structure building up from the molecular fragments is far more powerful.
1.3
DEGREE OF UNSATURATION
Traditionally, the molecular formula of a compound was derived from elemental analysis and its molecular weight which was determined independently. The concept of the degree of unsaturation of an organic compound derives simply from the tetravalency of carbon. For a non-cyclic hydrocarbon (i.e. an alkane) the number of hydrogen atoms must be twice the number of carbon atoms plus two, any “deficiency” in the number of hydrogens must be due to the presence of unsaturation, i.e. double bonds, triple bonds or rings in the structure. The degree of unsaturation can be calculated from the molecular formula for all compounds containing C, H, N, O, S or the halogens. There are 3 basic steps in calculating the degree of unsaturation: Step 1 - take the molecular formula and replace all halogens by hydrogens Step 2 - omit all of the sulfur or oxygen atoms Step 3 - for each nitrogen, omit the nitrogen and omit one hydrogen
3
Chapter 1 Introduction
After these 3 steps, the molecular formula is reduced to CnHmand the degree of unsaturation is given by: Degree of Unsaturation = n - ~
+1
The degree of unsaturation indicates the number of n bonds or rings that the compound contains. For example, a compound whose molecular formula is C4H9NO 2 is reduced to C4H8 which gives a degree of unsaturation of 1 and this indicates that the molecule must have one n bond or one ring. Note that any compound that contains an aromatic ring always has a degree of unsaturation greater than or equal to 4 , since the aromatic ring contains a ring plus three n bonds. Conversely, if a compound has a degree of unsaturation greater than 4, one should suspect the possibility that the structure contains an aromatic ring.
1.4
CONNECTIVITY
Even if it were possible to identify sufficient structural elements in a molecule to account for the molecular formula, it may not be possible to deduce the structural formula from a knowledge of the structural elements alone. For example, it could be demonstrated that a substance of molecular formula C3H 50C1 contains the structural elements: -
ch 3 — Cl
\
/
c=o
— ch 2and this leaves two possible structures:
C H o-C -C H 2-C I d II
CHo-CH2- C - C I and
J
ii
O
0
1
2
Not only the presence of various structural elements, but also their juxtaposition, must be determined to establish the structure of a molecule. Fortunately, spectroscopy often gives valuable information concerning the connectivity of structural elements and in the above example it would be very easy to determine whether there is a
4
Chapter 1 Introduction
ketonic carbonyl group (as in 1) or an acid chloride (as in 2). In addition, it is possible to determine independently whether the methyl (-CH3) and methylene (-CH2-) groups are separated (as in 1) or adjacent (as in 2).
1.5
SENSITIVITY
Sensitivity is generally taken to signify the limits of detectability of a chromophore. Some methods (e.g. [H NMR) detect all chromophores accessible to them with equal sensitivity while in other techniques (e.g. UV) the range of sensitivity towards different chromophores spans many orders of magnitude. In terms of overall sensitivity, i.e. the amount of sample required, it is generally observed that: MS > UV > IR > >H NMR > >3C NMR but considerations of relative sensitivity toward different chromophores may be more important.
1.6
PRACTICAL CONSIDERATIONS
The 5 major spectroscopic methods (MS, UV, IR, ]H NMR and ,3C NMR) have become established as the principal tools for the determination of the structures of organic compounds, because between them they detect a wide variety of structural elements. The instrumentation and skills involved in the use of all five major spectroscopic methods are now widely spread, but the ease of obtaining and interpreting the data from each method under real laboratory conditions varies. In very general terms: (a)
While the cost of each type of instrumentation differs greatly (NMR instruments cost between $50,000 and several million dollars), as an overall guide, MS and NMR instruments are much more costly than UV and IR spectrometers. With increasing cost goes increasing difficulty in maintenance and the required operator expertise, thus compounding the total outlay.
(b)
In terms of ease o f usage for routine operation, most UV and IR instruments are comparatively straightforward. NMR Spectrometers are also common as “hands-on” instruments in most chemistry laboratories and the users require routine training and a degree of basic computer literacy. Similarly some Mass Spectrometers are now designed to be used by researchers as “hands-on” routine instruments. However, the more advanced NMR Spectrometers and most Mass
5
Chapter 1 Introduction
Spectrometers are sophisticated instruments that are usually operated and maintained by specialists. (c)
The scope of each spectroscopic method can be defined as the amount of useful information it provides. This is a function of the total amount of information obtainable and also how difficult the data are to interpret. The scope of each method varies from problem to problem, and each method has its aficionados and specialists, but the overall utility undoubtedly decreases in the order: NMR > MS > IR > UY with the combination of ’H and l3C NMR providing the most useful information.
(d)
The theoretical background needed for each method varies with the nature of the experiment, but the minimum overall amount of theory needed decreases in the order: NMR » MS > UV « IR
6
2 ULTRAVIOLET (UV) SPECTROSCOPY
2.1
BASIC INSTRUMENTATION
Basic instrumentation for both UV and IR spectroscopy consists of an energy source, a sample cell, a dispersing device (prism or grating) and a detector, arranged as schematically shown in Figure 2.1.
dispersing device or prism
radiation source
Figure 2.1
detector
sample
Schematic Representation of an IR or UV Spectrometer
The drive of the dispersing device is synchronised with the x-axis of the recorder or fed directly to a computer, so that this indicates the wavelength of radiation reaching the detector. The signal from the detector is transmitted to the y-axis of the recorder or to a computer and this indicates how much radiation is absorbed by the sample at any particular wavelength. Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
7
Chapter 2 Ultraviolet Spectroscopy
In practice, double-beam instruments are used where the absorption of a reference cell, containing only solvent, is subtracted from the absorption of the sample cell. Double beam instruments also cancel out absorption due to the atmosphere in the optical path as well as the solvent. The energy source must be appropriate for the wavelengths of radiation being scanned. The materials from which the dispersing device and the detector are constructed must be as transparent as possible to wavelengths being scanned. For UV measurements, the cells and optical components are typically made of quartz and ethanol, hexane, water or dioxane are usually chosen as solvents.
2.2
THE NATURE OF ULTRAVIOLET SPECTROSCOPY
The term "UV spectroscopy" generally refers to electronic transitions occurring in the region of the electromagnetic spectrum (X in the range 200-380 nm) accessible to standard UV spectrometers. Electronic transitions are also responsible for absorption in the visible region (approximately 380-800 nm) which is easily accessible instrumentally but of less importance in the solution of structural problems, because most organic compounds are colourless. An extensive region at wavelengths shorter than ~ 200 nm ("vacuum ultraviolet") also corresponds to electronic transitions, but this region is not readily accessible with standard instruments. UV spectra used for determination of structures are invariably obtained in solution.
2.3
QUANTITATIVE ASPECTS OF ULTRAVIOLET SPECTROSCOPY
The y-axis of a UV spectrum may be calibrated in terms of the intensity of transmitted light (i.e. percentage of transmission or absorption), as is shown in Figure 2.2, or it may be calibrated on a logarithmic scale i.e. in terms of absorbance (A) defined in Figure 2.2. Absorbance is proportional to concentration and path length (the Beer-Lambert Law). The intensity of absorption is usually expressed in terms of molar absorbance or the molar extinction coefficient (s) given by:
where M is the molecular weight, C the concentration (in grams per litre) and / is the path length through the sample in centimetres.
Chapter 2 Ultraviolet Spectroscopy
Figure 2.2
Definition of Absorbance (A)
UV absorption bands (Figure 2.2) are characterised by the wavelength of the absorption maximum (/-max) and s. The values of s associated with commonly encountered chromophores vary between 10 and 105. For convenience, extinction coefficients are usually tabulated as logio(s) as this gives numerical values which are easier to manage. The presence of small amounts of strongly absorbing impurities may lead to errors in the interpretation of UV data.
2.4
CLASSIFICATION OF UV ABSORPTION BANDS
UV absorption bands have fine structure due to the presence of vibrational sub-levels, but this is rarely observed in solution due to collisional broadening. As the transitions are associated with changes of electron orbitals, they are often described in terms of the orbitals involved, e.g. a -> a* 71 —> K* n —» 7i* n
a*
where n denotes a non-bonding orbital, the asterisk denotes an antibonding orbital and ct and n have the usual significance. Another method of classification uses the symbols: B
(for benzenoid)
E
(for ethylenic)
R
(for radical-like)
K
(for conjugated - from the German "konjugierte")
9
Chapter 2 Ultraviolet Spectroscopy
A molecule may give rise to more than one band in its UV spectrum, either because it contains more than one chromophore or because more than one transition of a single chromophore is observed. However, UV spectra typically contain far fewer features (bands) than IR, MS or NMR spectra and therefore have a lower information content. The ultraviolet spectrum of acetophenone in ethanol contains 3 easily observed bands:
logio(s) O C.
(nm) CH,
acetophenone
2.5
Assignment
244
12,600
4.1
7t -> 71*
K
280
1,600
3.2
7T
71*
B
317
60
1.8
n
71*
R
SPECIAL TERMS IN UV SPECTROSCOPY
Auxochromes (auxiliary chromophores) are groups which have little UV absorption by themselves, but which often have significant effects on the absorption (both Xmax and s) of a chromophore to which they are attached. Generally, auxochromes are atoms with one or more lone pairs e.g. -OH, -OR, -NR2, -halogen. If a structural change, such as the attachment of an auxochrome, leads to the absorption maximum being shifted to a longer wavelength, the phenomenon is termed a bathochromic shift. A shift towards shorter wavelength is called a hypsochromic shift.
2.6
IMPORTANT UV CHROMOPHORES
Most of the reliable and useful data is due to relatively strongly absorbing chromophores (s > 2 00 ) which are mainly indicative of conjugated or aromatic systems. Examples listed below encompass most of the commonly encountered effects.
Chapter 2 Ultraviolet Spectroscopy
(1)
Dienes and Polyenes
Extension of conjugation in a carbon chain is always associated with a pronounced shift towards longer wavelength, and usually towards greater intensity (Table 2.1).
Table 2.1
The Effect of Extended Conjugation on UV Absorption
Alkene
165
10,000
4.0
CH3-CH2-CH=CH-CH2-CH3 (trans)
184
10,000
4.0
c h 2=ch-ch=ch 2
217
20,000
4.3
CH3-CH=CH-CH=CH2 (trans)
224
23,000
4.4
CH2=CH-CH=CH-CH=CH2 (trans)
263
53,000
4.7
CH3-(CH=CH)5-CH3 (trans)
341
126,000
5.1
X
II CM X
CM
logio(e)
o
e
o
^max(nm)
When there are more than 8 conjugated double bonds, the absorption maximum of polyenes is such that they absorb light strongly in the visible region of the spectrum. Empirical rules (Woodward's Rules) of good predictive value are available to estimate the positions of the absorption maxima in conjugated alkenes and conjugated carbonyl compounds. The stereochemistry and the presence of substituents also influence UV absorption by the diene chromophore. For example:
V x = 214nm
^max = 2 5 3 n m
s = 16,000
s = 8,000
logio(s) = 4.2
logio(s) = 3.9
11
Chapter 2 Ultraviolet Spectroscopy
(2)
Carbonyl compounds
All carbonyl derivatives exhibit weak (s < 100) absorption between 250 and 350 nm, and this is only of marginal use in determining structure. However, conjugated carbonyl derivatives always exhibit strong absorption (Table 2.2).
Table 2.2
UV Absorption Bands in Common Carbonyl Compounds
Compound
o
/
w
CO
X
o -x
Acetone
o
Acetaldehyde
Structure
C H 3 x
^0 c 1 ch3
c
c
1
1
CH3
4-Methylpent-3-en-2-one ch3
Cyclohex-2-en-1 -one
Benzoquinone
12
1
ch3
o=° 0 = = o
Sulfones
i o=s=o I
Sulfonamides and Sulfonate esters
— SOo-N^
Alcohols
— C-OH
^
\
— s o 2- o —
/
Ethers Alkyl fluorides
\
— C-OR /
\ — C -F
|
1
/
Alkyl chlorides Alkyl bromides Alkyl iodides
\
— C -C I / \
19
Chapter 3 Infrared Spectroscopy
Carbon-carbon double bonds in unconjugated alkenes usually exhibit weak to moderate absorptions due to C=C stretching in the range 1660-1640 cm'1. Disubstituted, trisubstituted and tetrasubstituted alkenes usually absorb near 1670 cm'1. The more polar carbon-carbon double bonds in enol ethers and enones usually absorb strongly between 1600 and 1700 cm-1. Alkenes conjugated with an aromatic ring absorb strongly near 1625 cm-1. (4)
Chromophores absorbing in the region between 1900 and 2600 cm-1- The
absorptions listed in Table 3.3 often yield useful information because, even though some are of only weak or medium intensity, they occur in regions largely devoid of absorption by other commonly occurring chromophores.
20
Functional group
Structure
v (cm '1)
Intensity
alkyne
— c= c—
2100-2300
weak to medium
nitrile
---- C = N
2215 -2280
medium
cyanate
---- O— C = N
2130-2270
strong
thiocyanate
---- S— C = N
2130-2175
medium
isocyanate
---- N = C — O
2200 - 2300
strong broad
isothiocyanate
---- N = C = S
2000 - 2200
strong
allene
II o
Table 3.3 Common IR Absorption Frequencies in the Region 1900 - 2600 cm'1
1900 - 2000
strong
4 MASS SPECTROMETRY
It is possible to determine the masses of individual ions in the gas phase. Strictly speaking, it is only possible to measure their mass/charge ratio (m/e), but as multi charged ions are very much less abundant than those with a single electronic charge (e= 1), m/e is for all practical purposes equal to the mass of the ion, m. The principal experimental problems in mass spectrometry are firstly to volatilise the substrate (which implies high vacuum) and secondly to ionise the neutral molecules to charged species.
4.1
IONISATION PROCESSES
The most common method of ionisation involves Electron Impact (El) and there are two general courses of events following a collision of a molecule M with an electron e. By far the most probable event involves electron ejection which yields an odd-electron positively charged cation radical [M]+‘ of the same mass as the initial molecule M. M + e —►
[M]+- + 2e
The cation radical produced is known as the molecular ion and its mass gives a direct measure of the molecular weight of a substance. An alternative, far less probable process, also takes place and it involves the capture of an electron to give a negative anion radical, [M]-. M + e —»
[M]"
Electron impact mass spectrometers are generally set up to detect only positive ions, but negative-ion mass spectrometry is also possible. The energy of the electron responsible for the ionisation process can be varied. It must be sufficient to knock out an electron and this threshold, typically about 10-12 eV, is known as the appearance potential. In practice much higher energies (~70 eV) are used and this large excess energy (1 eV = 95 kJ moh1) causes further fragmentation of the molecular ion. Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
21
Chapter 4 Mass Spectrometry
The two important types of fragmentation are: [M]+‘ —»
A+ (even electron cation) + B' (radical)
[M]+- —>
C+’ (cation radical) + D (neutral molecule)
or
As only species bearing a positive charge will be detected, the mass spectrum will show signals due not only to [M]+- but also due to A+, C+- and to fragment ions resulting from subsequent fragmentation of A+ and C+\ As any species may fragment in a variety of ways, the typical mass spectrum consists of many signals. The mass spectrum consists of a plot of masses of ions against their relative abundance. There are a number of other methods for ionising the sample in a mass spectrometer. The most important alternative ionisation method to electron impact is Chemical Ionisation (Cl). In Cl mass spectrometry, an intermediate substance (generally methane or ammonia) is introduced at a higher concentration than that of the substance being investigated. The carrier gas is ionised by electron impact and the substrate is then ionised by collisions with these ions. Cl is a milder ionisation method than El and leads to less fragmentation of the molecular ion. Another common method of ionisation is Electrospray Ionisation (ESI). In this method, the sample is dissolved in a polar, volatile solvent and pumped through a fine metal nozzle, the tip of which is charged with a high voltage. This produces charged droplets from which the solvent rapidly evaporates to leave naked ions which pass into the mass spectrometer. ESI is also a relatively mild form of ionisation and is very suitable for biological samples which are usually quite soluble in polar solvents but which are relatively difficult to vaporise in the solid state. Electrospray ionisation tends to lead to less fragmentation of the molecular ion than EL Matrix Assisted Laser Desorption Ionisation (M ALDI) uses a pulse of laser light to bring about ionisation. The sample is usually mixed with a highly absorbing compound which acts as a supporting matrix. The laser pulse ionises and vaporises the matrix and the sample to give ions which pass into the mass spectrometer. Again MALDI is a relatively mild form of ionisation which tends to give less fragmentation of the molecular ion than El. All of the subsequent discussion of mass spectrometry is limited to positive-ion electron-impact mass spectrometry.
22
Chapter 4 Mass Spectrometry
4.2
INSTRUMENTATION
In a magnetic sector mass spectrometer (Figure 4.1), the positively charged ions of mass, m, and charge, e (generally e= 1) are subjected to an accelerating voltage V and passed through a magnetic field H which causes them to be deflected into a curved path of radius r. The quantities are connected by the relationship: m
H2r 2
e
2V
The values of FI and V are known, r is determined experimentally and e is assumed to be unity thus permitting us to determine the mass m. In practice the magnetic field is scanned so that streams of ions of different mass pass sequentially to the detecting system (ion collector). The whole system (Figure 4.1) is under high vacuum (less than 10'6 Torr) to permit the volatilisation of the sample and so that the passage of ions is not impeded. The introduction of the sample into the ion chamber at high vacuum requires a complex sample inlet system.
High vacuum
Figure 4.1
Large negative potential
Magnetic Field
Schematic Diagram of an Electron-Impact Mass Spectrometer
23
Chapter 4 Mass Spectrometry
The magnetic scan is synchronised with the x-axis of a recorder and calibrated to appear as mass number (strictly m/e). The amplified current from the ion collector gives the relative abundance of ions on the y-axis. The signals are usually preprocessed by a computer that assigns a relative abundance of 100 % to the strongest peak (base peak).
Many modem mass spectrometers do not use a magnet to bend the ion beam to separate ions but rather use the “time o f flight” (TOF) of an ion over a fixed distance to measure its mass. In these spectrometers, ions are generated (usually using a very short laser pulse) then accelerated in an electric field. Lighter ions have a higher velocity as they leave the accelerating field and their time of flight over a fixed distance will vary depending on the speed that they are travelling. Time of Flight mass spectrometers have the advantage that they do not require large, high-precision magnets to bend and disperse the ion beam so they tend to be much smaller, compact and less complex (desk-top size) instruments.
4.3
MASS SPECTRAL DATA
As well as giving the molecular weight of a substance, the molecular ion of a compound may provide additional information. The "nitrogen rule" states that a molecule with an even molecular weight must contain no nitrogen atoms or an even number of nitrogen atoms. This means that a molecule with an odd molecular weight must contain an odd number of nitrogen atoms. (1)
High resolution mass spectra. The mass of an ion is routinely determined to
the nearest unit value. Thus the mass of [M]+' gives a direct measure of molecular weight. It is not usually possible to assign a molecular formula to a compound on the basis of the integer m/e value of its parent ion. For example, a parent ion at m/e 72 could be due to a compound whose molecular formula is C4H80 or one with a molecular formula C3H40 2 or one with a molecular formula C3H 8N2. However, using a double-focussing mass spectrometer or a time-of-flight mass spectrometer, the mass of an ion or any fragment can be determined to an accuracy of approximately ± 0.00001 of a mass unit (a high resolution mass spectrum). Since the masses of the atoms of each element are known to high accuracy, molecules that may have the same mass when measured only to the nearest integer mass unit, can be distinguished when the mass is measured with high precision. Based on the accurate masses of 12C, 160 , 14N and 'H (Table 4.1) ions with the formulas C4HgO+', C 3H40 2+’ or C3HgN2+’ would have accurate masses 72.0573, 72.0210, and 72.0686 so these
24
Chapter 4 Mass Spectrometry
could easily be distinguished by high resolution mass spectroscopy. In general, if the mass of any fragment in the mass spectrum can be accurately determined, there is usually only one combination of elements which can give rise to that signal since there are only a limited number of elements and their masses are accurately known. By examining a mass spectrum at sufficiently high resolution, one can obtain the exact composition of each ion in a mass spectrum, unambiguously. Most importantly, determining the accurate mass of [M]+' gives the molecular formula o f the compound.
Table 4.1
Isotope
Accurate Masses of Selected Isotopes
Natural
Mass
Abundance (%)
•H
99.98
1.00783
2H 12C
0.016
2.01410
98.9
12.0000
13C
1.1
13.00336
14N
99.6
14.0031
15N 16Q
0.37
15.0001
99.8
15.9949
17Q
0.037
16.9991
18Q 19p
0.20
17.9992
100
18.99840
28Si
92.28
27.9769
29Si
4.7
28.9765
30Si 31P 32$
3.02
29.9738
100
30.97376
95.0
31.9721
33S
0.75
32.9715
34S
4.2
33.9679
35C1
75.8
34.9689
37C1
24.2
36.9659
79Br
50.7
78.9183
81Br
49.3
80.9163
127I
100
126.9045
25
Chapter 4 Mass Spectrometry
(2)
Molecular Fragmentation. The fragmentation pattern is a molecular
fingerprint. In addition to the molecular ion peak, the mass spectrum (see Figure 4.1) consists of a number of peaks at lower mass numbers and these result from fragmentation of the molecular ion. The principles determining the mode of fragmentation are reasonably well understood, and it is possible to derive structural information from the fragmentation pattern in several ways. (a)
The appearance of prominent peaks at certain mass numbers can be correlated empirically with certain structural elements (Table 4.2), e.g. a prominent peak at m/e = 43 is a strong indication of the presence of a CH3-CO- group in the molecule.
(b)
Information can also be obtained from differences between the masses of two peaks. Thus a prominent fragment ion that occurs 15 mass numbers below the molecular ion, suggests strongly the loss of a CH3- group and therefore that a methyl group was present in the substance examined.
(c)
The knowledge of the principles governing the mode of fragmentation of ions makes it possible to confirm the structure assigned to a compound and, quite often, to determine the juxtaposition of structural fragments and to distinguish between isomeric substances. For example, the mass spectrum of benzyl methyl ketone, Ph-CH2-CO-CH3 contains a strong peak at m/e = 91 due to the stable ion Ph-CH2+, but this ion is absent in the mass spectrum of the isomeric propiophenone Ph-CO-CH2CH 3 where the structural elements Ph- and -CH2- are separated. Instead, a prominent peak occurs at m/e = 105 due to the stable ion Ph-C=0+.
Electronic databases of the mass spectral fragmentation patterns of known molecules can be rapidly searched by computer. The pattern and intensity of fragments in the mass spectrum is characteristic of an individual compound so comparison of the experimental mass spectrum of a compound with those in a library can be used to positively identify it, if its spectrum has been recorded previously.
26
Chapter 4 Mass Spectrometry
Table 4.2
Common Fragments and their Masses
Fragment
Mass
Fragment
Mass
Fragment
Mass
O II
CH3-
15
CH3CH2-
29
NO
30
— CH2OH
31
/C —
29
CH2=CH-CH2
41
H
O II
o II
43
.c — ch
HO
—N 02
45
46
3^ o
C 4h7
.OH
55
C 4H9
57
60
C 5H5
65
CH3CH2^
/
_
\ _
57
77
ch 2= c
X° H
c 6h 5
_
91
xx - f u
N O ^ c h 2-
1%) of minor isotopes, is often obvious simply by inspection of ions near the molecular ion.
27
Chapter 4 Mass Spectrometry
The relative intensities of the [M]+‘, [M+l]+‘ and [M+2]+‘ ions exhibit a characteristic pattern depending on the specific isotopes that make up the ion. For any molecular ion (or fragment) which contains one bromine atom, the mass spectrum will contain two peaks separated by two m/e units, one for the ions which contain 79Br and one for the ions which contain 81Br. For bromine-containing fragments, the relative intensities of the two ions will be approximately the same, since the natural abundances of 79Br and 8lBr are 100
approximately equal. In the mass spectrum of 1-bromoethane, there are two molecular ions of almost equal intensity. The peak at m/e 108 corresponds to the molecular ions in the sample which contain 79Br; the peak at
Mass Spectrum
60 40
C 2H 581B r+ '
CO - 0 - CL - (D ■ 0 - CO . -Q 7 o “ as
M = 108 C 2H 579Br+ '
20 7 u li ____ ___i__i__i__i__i__i__i__i__i__.__i---------------
40
m/e 110 corresponds to the molecular ions in the sample which contain
81
M = 110
C 2H 5Br
80 -•
80 m /e
120
Br.
Similarly, for any molecule (or fragment of an molecule) which contains one chlorine atom, the mass spectrum will contain two fragments separated by two m/e units, one for the ions which contain 35Cl and one for the ions which contain 37C1. For chlorinecontaining ions, the relative intensities of the two ions will be approximately 3:1 since this reflects the natural abundances of 35C1 and 37C1. In the mass spectrum of 2-chloropropane, there are two molecular ions at 78 and 80
100 -
Mass Spectrum 43
with intensities approximately in the ratio
80 -•
3:1. The peak at m/e 78 corresponds to the
60 »
molecular ions in the sample which contain
40
35C1; the peak at m/e 80 corresponds to the
20
CO
- 0 . CL - 0 .
M = 78
0 CO
c
-Q 7 o
M = 80 c ^ 3, nh
molecular ions in the sample which contain 37C1. Note that the base peak at m/e 43 is
only a single ion so this ion must contain no
35^.+' ci
3h / °
40
^ f 7737c i + ‘ I I I—a L 120 80
m /e
chlorine. The pair of ions at m/e 63 and 65 clearly corresponds to a fragment that still contains a chlorine atom. Any molecular ion (or fragment) which contains 2 bromine atoms will have a pattern of ions M:M+2:M+4 with signals in the ratio 1:2:1 and any molecular ion (or fragment) which contains 2 chlorine atoms will have a pattern of M:M+2:M+4 with signals in the ratio 10:6.5:1 (Figure 4.2).
28
Chapter 4 Mass Spectrometry
Br
Brc CM +
O MT O CD
Figure 4.2
+
T- oo ^ r to
Cl
B ra CM NT CD + + +
CO O
CD CO
Cl2
CM
+
O o
vCO
O LOo O
CD t —
Cl3
BrCI
CM NT CD + + +
CM
O LO-r- CO OCJ) CO
OOOMN O CM
+ +
BrCI2 CM
CD
+++
Br2CI CM
CD
+++
CM O L O CD CD O M -
Relative Intensities of the Cluster of Molecular Ions for Molecules Containing Combinations of Bromine and Chlorine Atoms
(4)
Chromatography coupled with Mass Spectrometry. It is now common to
couple an instrument for separating a mixture of organic compounds e.g. using gas chromatography (GC) or high performance liquid chromatography (HPLC), directly to the input of a mass spectrometer. In this way, as each individual compound is separated from the mixture, its mass spectrum can be recorded and compared automatically with the library of known compounds and identified immediately if it is a known compound. (5)
Metastable peaks in a mass spectrum arise if the fragmentation process a+ —> b+ + c (neutral)
takes place within the ion-accelerating region of the mass spectrometer (Figure 4.1). Ion peaks corresponding to the masses of a+ and to b+ (ma and /??/,) may be accompanied by a broader peak at mass m , such that: 2
mb m* = ------ma
The presence of metastable peaks in a mass spectrum often permits positive identification of a particular fragmentation path.
4.4
REPRESENTATION OF FRAGMENTATION PROCESSES
As fragmentation reactions in a mass spectrometer involve the breaking of bonds, they can be represented by the standard "arrow notation" used in organic chemistry. For some purposes a radical cation (e.g. a generalised ion of the molecular ion) can be represented without attempting to localise the missing electron: [M]+- or
[H3C-CH2-0-R] +•
29
Chapter 4 Mass Spectrometry
However, to show a fragmentation process it is generally necessary to indicate "from where the electron is missing" even though no information about this exists. In the case of the molecular ion corresponding to an alkyl ethyl ether, it can be reasonably inferred that the missing electron resided on the oxygen. The application of standard arrow notation permits us to represent a commonly observed process, viz. the loss of a methyl fragment from the [H3C-CH2-0 -R ]+' molecular ion:
4.5
FACTORS GOVERNING FRAGMENTATION PROCESSES
Three factors dominate the fragmentation processes: (a)
Weak bonds tend to be broken most easily
(b)
Stable fragments (not only ions, but also the accompanying radicals and molecules) tend to be formed most readily
(c)
Some fragmentation processes depend on the ability of molecules to assume cyclic transition states.
Favourable fragmentation processes naturally occur more often and ions thus formed give rise to strong peaks in the mass spectrum.
4.6
EXAMPLES OF COMMON TYPES OF FRAGMENTATION
There are a number of common types of cleavage which are characteristic of various classes of organic compounds. These result in the loss of well-defined fragments which are characteristic of certain functional groups or structural elements. (1)
Cleavage at Branch Points. Cleavage of aliphatic carbon skeletons at branch
points is favoured as it leads to more substituted (and hence more stable) carbocations. The mass spectrum of 2,2-dimethylpentane shows strong peaks at m/e = 85 and m/e = 57 where cleavage leads to the formation of stable tertiary carbocations.
30
Chapter 4 Mass Spectrometry
CH3
ch3
I
+• I
CH3— C—CHo—CHo—c h 3
c h 3—c —c h 2—c h 2—c h 3
cm
ch3
m/e = 100
ch3
CH, ch3
c h 3—c —c h 2- c h 2- c h 3 ch3
C—CH2—CH2—CH3 ch3
neutral fragment
stable cation
m/e = 85 ch3 I+
CH3 +• I c h 3—c —c h 2—c h 2—c h 3
cm—c
CH
ch
CH,
stable cation
2— c h 2— c h 3
neutral fragment
m/e = 57 - Cleavage. Chain cleavage tends to occur P to heteroatoms, double bonds and aromatic rings because relatively stable, delocalised carbocations result in each case. (a) .. I I R - X — C — C— I I
X=
R - X — C— CI I
0 , N, S, halogen + / R -X =C n
+ / R -X -C n
\
resonance stabilised carbocation
(b)
V - r / / C- ° \
neutral fragment
' + I
c— c-
x l
-e-
/
.C-C
I
W
\ /
/
C— C \
. +
s +
c—
« t C -C *— ► /
/
I
l
c— I
resonance stabilised carbocation
. / C—
\
neutral fragment
Chapter 4 Mass Spectrometry
(c)
\C — C I — I
a
- e -
I
resonance stabilised carbocation
(3)
Cleavage a to carbonyl groups. Cleavage tends to occur a to carbonyl
groups to give stable acylium cations. R may be an alkyl, -OH or -OR group.
.C
^
^
R /
c C \
^
c R
I ^ a
resonance stabilised carbocation
/
*c— \ neutral fragment
O II C + I R
(4)
Cleavage a to heteroatoms. Cleavage of chains may also occur a to
heteroatoms, e.g. in the case of ethers:
I R -O -C " I
(5)
. R -O -C " |
► R -0: C" I free radical carbocation
Retro Diels-Alder reaction. Cyclohexene derivatives may undergo a retro
Diels-Alder reaction:
32
-e'
Chapter 4 Mass Spectrometry
(6)
The McLafferty rearrangement. Compounds where the molecular ion can
assume the appropriate 6 -membered cyclic transition state usually undergo a cyclic fragmentation, known as the McLafferty rearrangement. This rearrangement involves a transfer of a y hydrogen atom to an oxygen and is often observed with ketones, acids and esters:
With primary carboxylic acids, R-CHr COOH, this fragmentation leads to a characteristic peak at m/e = 60 +
h2c = c — oh
OH With carboxylic esters, two types of McLafferty rearrangements may be observed and ions resulting from either fragmentation pathway are commonly observed in the mass spectrum:
5 NUCLEAR MAGNETIC RESONANCE (NMR) SPECTROSCOPY
5.1
THE PHYSICS OF NUCLEAR SPINS AND NMR INSTRUMENTS
(1)
The Larmor Equation and Nuclear Magnetic Resonance
All nuclei have charge because they contain protons and some of them also behave as if they spin. A spinning charge generates a magnetic dipole and is associated with a small magnetic field H (Figure 5.1). Such nuclear magnetic dipoles are characterised by nuclear magnetic spin quantum numbers which are designated by the letter I and can take up values equal to 0 , V2, 1 , 3/2 ... etc.
H
Figure 5.1
A Spinning Positive Charge Generates a Magnetic Field and Behaves like a Small Magnet
It is useful to consider three types of nuclei: Type 1:
Nuclei with 1 = 0. These nuclei do not interact with the applied magnetic field and are not NMR chromophores. Nuclei with 1 = 0 have an even number of protons and even number of neutrons and have no net spin. This means that nuclear spin is a property characteristic of certain isotopes rather than of certain elements. The most prominent examples of nuclei with I = 0 are !2C and 160 , the dominant isotopes of carbon and oxygen. Both oxygen and carbon also have isotopes that can be observed by NMR spectroscopy.
34
Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
Chapter 5 NMR Spectroscopy
Type 2:
Nuclei with I = V2. These nuclei have a non-zero magnetic moment and are NMR visible and have no nuclear electric quadrupole (Q). The two most important nuclei for NMR spectroscopy belong to this category: *H (ordinary hydrogen) and 13C (a non-radioactive isotope of carbon occurring to the extent of 1.06% at natural abundance). Also, two other commonly observed nuclei 19F and 31P have I = A Together, NMR data for 'H and 13C account for well over 90% of all NMR observations in the literature and the discussion and examples in this book mostly refer to these two nuclei. However, the spectra of all nuclei with I = x/i can be understood easily on the basis of common theory.
Type 3:
Nuclei with I > ‘A These nuclei have both a magnetic moment and an electric quadrupole. This group includes some common isotopes (e.g. 2H and 14N) but they are more difficult to observe and spectra are generally very broad. This group of nuclei will not be discussed further.
The most important consequence of nuclear spin is that in a uniform magnetic field, a nucleus of spin I may assume 21 + 1 orientations. For nuclei with I = A there are just 2 permissible orientations (since 2 x y2 + 1 = 2). These two orientations will be of unequal energy (by analogy with the parallel and antiparallel orientations of a bar magnet in a magnetic field) and it is possible to induce a spectroscopic transition (spin-flip) by the absorption of a quantum of electromagnetic energy (AE) of the appropriate frequency (v):
h
(5-1)
In the case of NMR, the energy required to induce the nuclear spin flip also depends on the strength of the applied field, H 0. It is found that: v = KHq
(5.2)
where ATis a constant characteristic of the nucleus observed. Equation 5.2 is known as the Larmor equation and is the fundamental relationship in NMR spectroscopy. Unlike other forms of spectroscopy, in NMR the frequency of the absorbed electromagnetic radiation is not anabsolute value for anyparticular transition, but has a different value depending on thestrength of the applied magnetic field. For every value of H 0, there is a matching value of v corresponding to the condition of resonance according to Equation 5.2, and this is the origin of the term “resonance ” in Nuclear Magnetic Resonance Spectroscopy. Thus for 'H and 13C, resonance
35
Chapter 5 NMR Spectroscopy
frequencies corresponding to magnitudes of applied magnetic field (H 0) commonly found in commercial instruments are given in Table 5.1.
Table 5.1
Resonance Frequencies of *H and 13C Nuclei in Magnetic Fields of Different Strengths
v *H (MHz)
v 13C (MHz)
H 0 (Tesla)
60
15.087
1.4093
90
22.629
2.1139
100
25.144
2.3488
200
50.288
4.6975
300
75.432
7.0462
400
100.577
9.3950
500
125.720
11.744
600
150.864
14.0923
750
188.580
17.616
800
201.154
18.790
900
226.296
21.128
In common jargon, NMR spectrometers are commonly known by the frequency they use to observe ’H i.e. as "60 MHz", "200 MHz" or "400 MHz" instruments, even if the spectrometer is set to observe a nucleus other than 'H. All the frequencies listed in Table 5.1 correspond to the radio frequency region of the electromagnetic spectrum and inserting these values into Equation 5.1 gives the size of the energy gap between the states in an NMR experiment. A resonance frequency of 100 MHz corresponds to an energy gap of approximately 4 x 10’5 kJ mob1. This is an extremely small value on the chemical energy scale and this means that NMR spectroscopy is, for all practical purposes, a ground-state phenomenon. Any absorption signal observed in a spectroscopic experiment must originate from excess of the population in the lower energy state, the so called Boltzmann excess, which is equal to N^-Na, where iVp and Na are the populations in the lower ((3) and upper (a) energy states.
36
Chapter 5 NMR Spectroscopy
For molar quantities, the general Boltzmann relation (Equation 5.3) shows that: AE iV p
RT
=
e
(5 J )
N H —
area = 3
1 : 2:1
-CHdoublet area = 3
1 :3 :3 : 1
1:1
—CH2—
-C H 2-
triplet
triplet
area = 2
area = 2
1 : 2:1
1 : 2:1
-(CH3)2
/CH3
doublet
—CH
area = 6
xCH3
an isopropyl grc
1 : 6 : 15 : 20 : 15 : 6 : 1
1:1
55
Chapter 5 NMR Spectroscopy
5.7
ANALYSIS OF 1H NMR SPECTRA
To obtain structurally useful information from NMR spectra, one must solve two separate problems. Firstly, the spectrum must be analysed to obtain the NMR parameters (chemical shifts and coupling constants) for all the protons and, secondly, the values of the coupling constants must be interpreted in terms of established relationships between the parameters and structure. (1)
A spin system is defined as a group of coupled protons. Clearly, a spin system
cannot extend beyond the bounds of a molecule, but it may not include a whole molecule. For example, isopropyl propionate comprises two separate and isolated proton spin systems, a seven-proton system for the isopropyl residue and a five-proton system for the propionate residue, because the ester group effectively provides a barrier (5 bonds) against coupling between the two parts.
c h 3x CH O -C CH3x 7-spin system
(2)
ii
CH2CH3 Isopropyl propionate
O 5-spin system
Strongly and weakly coupled spins. These terms refer not to the actual
magnitude of J, but to the ratio of the separation of chemical shifts expressed in Hz (Av) to the coupling constant J between them. For most purposes, if Av/J is larger than ~3, the spin system is termed weakly coupled. When this ratio is smaller than ~3, the spins are termed strongly coupled. Two important conclusions follow: (a)
Because the chemical shift separation (Av) is expressed in Hz, rather than in the dimensionless 8 units, its value will change with the operating frequency of the spectrometer, whiie the value of J remains constant. It follows that two spins will become progressively more weakly coupled as the spectrometer frequency increases. Weakly coupled spin systems are much easier to analyse than strongly coupled spin systems and thus spectrometers operating at higher frequencies (and therefore at higher applied magnetic fields) will yield spectra which are more easily interpreted. This has been an important reason for the development of NMR spectrometers operating at ever higher magnetic fields.
56
Chapter 5 NMR Spectroscopy
(b)
Within a spin system, some pairs of nuclei or groups of nuclei may be strongly coupled and others weakly coupled. Thus a spin-system may be partially strongly coupled.
(3)
Magnetic equivalence. A group of protons is magnetically equivalent when
they not only have the same chemical shift (chemical equivalence) but also have identical spin-spin coupling to each individual nucleus outside the group. (4)
Conventions used in naming spin systems. Consecutive letters of the
alphabet (e.g. A, B, C D ,
) are used to describe groups of protons which are
strongly coupled. Subscripts are used to give the number of protons that are magnetically equivalent. Primes are used to denote protons that are chemically equivalent but not magnetically equivalent. A break in the alphabet indicates weakly coupled groups. For example: ABC denotes a strongly coupled 3-spin system AMX denotes a weakly coupled 3-spin system ABX denotes a partially strongly coupled 3-spin system A3BMXY denotes a spin system in which the three magnetically equivalent A nuclei are strongly coupled to the B nucleus, but weakly coupled to the M, X and Y nuclei. The nucleus X is strongly coupled to the nucleus Y but weakly coupled to all the other nuclei. The nucleus M is weakly coupled to all the other 6 nuclei. AA'XX' is a 4-spin system described by two chemical shift parameters (for the nuclei A and X) but where JAX ^ JAX,. A and A' (as well as X and X') are pairs of nuclei which are chemically equivalent but magnetically non-equivalent. The process of deriving the NMR parameters (8 and J) from a set of multiplets in a spin system is known as the analysis o f the NMR spectrum. In principle, any spectrum arising from a spin system, however complicated, can be analysed but some will require calculations or simulations performed by a computer. Fortunately, in a very large number of cases, multiplets can be correctly analysed by inspection and direct measurements. These spectra are known as first order spectra and they arise from weakly coupled spin systems. At high applied magnetic fields, a large proportion of 'H NMR spectra are nearly pure first-order and there is a tendency for simple molecules, e.g. those exemplified in the problems in this text, to exhibit first-order spectra even at moderate fields.
57
Chapter 5 NMR Spectroscopy
5.8
CHANGING THE MAGNETIC FIELD IN NMR SPECTROSCOPY
Changing the magnetic field of an NMR spectrometer i.e. recording the NMR spectrum of a sample on NMR instruments operating at different magnetic fields, highlights several important aspects of NMR spectroscopy. The chemical shifts of the nuclei (expressed in Hz) are proportional to the strength of the magnetic field (see Section 5.1). The higher the magnetic field, the higher the resonance frequency of each nucleus in the sample. Nuclei move further apart (in Hz) from each other as the magnetic field strength increases i.e. there is better dispersion in the spectrum. On the other hand, spin-spin coupling is a molecular property that is independent of the magnetic field of the spectrometer. So the splittings in an NMR spectrum due to spin-spin coupling, remain constant, irrespective of the magnetic field strength of the spectrometer. Given below are H NMR spectra of 2-bromotoluene where spectra on the same sample have been recorded in 3 different NMR spectrometers each operating at a different magnetic field strength. There are 4 different aromatic proton resonances. At the lowest field (4.7 Tesla), there is significant overlap of the resonances. The resonances for H3 and H 6 are severely distorted and this is not a 1st order spectrum. At 14.1 Tesla (600 MHz) dispersion is much better and the spectrum is a 1st order spectrum which could be analysed by 1st order rules. 600 MHz spectrometer 14.1 Tesla magnet
Aromatic region of the 'H NMR spectrum of 2 -bromotoluene
400 MHz spectrometer 9.4 Tesla magnet
(acetone-^, solution).
200 MHz spectrometer 4.7 Tesla magnet
Chapter 5 NMR Spectroscopy
5.9
Rule 1
RULES FOR SPECTRAL ANALYSIS OF FIRST ORDER SPECTRA
A group of n magnetically equivalent protons will split a resonance of an interacting group of protons into n + 1 lines. For example, the resonance due to the A protons in an A nX m system will be split into m+ 1 lines, while the resonance due to the X protons will be split into n+ 1 lines. More generally, splitting by n nuclei of spin quantum number I, results in 2 «I +1 lines. This simply reduces to n+1 for protons where I = V2.
Rule 2
The spacing (measured in Hz) of the lines in the multiplet will be equal to the coupling constant. In the above example, all spacings in both parts of the spectrum will be equal to JAX.
Rule 3
The true chemical shift of each group of interacting protons lies in the centre of the (always symmetrical) multiplet.
Rule 4
The relative intensities of the lines within each multiplet will be in the ratio of the binomial coefficients (Table 5.9). Note that, in the case of higher multiplets, the outside components of multiplets are relatively weak and may be lost in the instrumental noise, e.g. a septet may appear as a quintet if the outer lines are not clearly visible. The intensity relationship is the first to be significantly distorted in non-ideal cases, but this does not lead to serious errors in spectral analysis.
Rule 5
When a group of magnetically equivalent protons interacts with more than one group of protons, its resonance will take the form of a multiplet o f multiplets. For example, the resonance due to the A protons in a system AW M XOTwill have the multiplicity of (p+X){m+ \). The multiplet patterns are chained e.g. a proton coupled to 2 different protons will be split to a doublet by coupling to the first proton then each of the component of the doublet will be split further by coupling to the second proton resulting in a symmetrical multiplet with 4 lines (a doublet of doublets).
59
Chapter 5 NMR Spectroscopy
Ha U
LI
yM | yx X -C -C -C -Y doublet of doublets
Ha |_| J yM | L yx X -C -C -C -Y HM 1 1
triplet of doublets or doublet of triplets
The appropriate coupling constants will control splitting and relative intensities will obey rule 4. Rule
6
Rule 7
Protons that are magnetically equivalent do not split each other. Any system A„ will give rise to a singlet. Spin systems that contain groups of chemically equivalent protons that are not magnetically equivalent cannot be analysed by first-order methods.
Rule
8
If Avab/Tab is less than ~3, for any pair of nuclei A and B in the spin system, the spectra become distorted from the expected ideal multiplet patterns and the spectra cannot be analysed by first-order methods.
60
Chapter 5 NMR Spectroscopy
(1) S p littin g D ia g ra m s
The knowledge of the rules listed above, permits the development of a simple procedure for the analysis of any spectrum which is suspected of being first order. The first step consists of drawing a splitting diagram, from which the line spacings can be measured and identical (hence related) splittings can be identified (Figure 5.8).
8 (ppm from TMS)
Figure 5.8
A Portion of the *H NMR Spectrum of Styrene Epoxide (100 MHz as a 5% solution in CCI4)
The section of the spectrum of styrene epoxide (Figure 5.8) clearly contains the signals from 3 separate protons (identified as Hi, H 2 and H3) with Hi at 5 2.95, H2 at 5 2.58 and H 3 at 5 3.67 ppm. Each signal appears as a doublet of doublets and the chemical shift of each proton is simply obtained by locating the centre of each multiplet. The pair of nuclei giving rise to each splitting is clearly indicated by the splitting diagram above each multiplet with 2J H,_H2 = 5.9 Hz, V H1.H3 = 4.0 Hz and = 2.5 Hz. The validity of a first order analysis can be verified by calculating the ratio Av/J for each pair of nuclei and establishing that it is greater than 3 .
61
Chapter 5 NMR Spectroscopy
From Figure 5.8 ^
.
IL
.
5.9
J-\ 2
6.3
^13 =
72
J-13
4.0
= ig o
Av23 _ 109 = 43 6
J 23 2.5
Each ratio is greater than 3 soa first order analysis is justified and the 100 MHz spectrum of the aliphatic protons of styrene oxide is indeeda first order spectrum and could be labelled as an AMX spin system. The 60 MHz ’H spectrum of a 4 spin AMX2 system is given in Figure 5.9. This system contains 3 separate proton signals (in the intensity ratios 1 : 1 : 2 , identified as Ha, Hm and Hx). The multiplicity of HA is a triplet of doublets, the multiplicity of HM is a triplet of doublets and the multiplicity of Hx is a doublet of doublets. Again, the nuclei giving rise to each splitting are clearly indicated by the splitting diagram above each multiplet and the chemical shifts of each multiplet are simply obtained by measuring the centres of each multiplet.
Hx 8 = 7.0 ppm
8 = 6.0 ppm
8 = 4.95 ppm
v = 420 Hz
v = 360 Hz
v = 297 Hz
7.0
Figure 5.9
62
6.0
5.0
8 (ppm from TMS)
The 60 MHz 'H NMR Spectrum of a 4-Spin AMX2 Spin System
Chapter 5 NMR Spectroscopy
A spin system comprising just two protons (i.e. an AX or an AB system) is always exceptionally easy to analyse because, independent of the value of the ratio of Av/J, the spectrum always consists of just four lines with each pair of lines separated by the coupling constant J. The only distortion from the first-order pattern consists of the gradual reduction of intensities of the outer lines in favour of the inner lines, a characteristic "sloping" or "tenting" towards the coupling partner. A series of simulated spectra of two-spin systems are shown in Figure 5.10.
12
Av12
—
J
Ul
= 10.0
J12
Av.12
3.0
12
uu Av12 — 12 = 0.8 J12 jU ul
Av12 — 2 = 0.0 J 12
-j----- 1------- 1----1-------1---- 1----- 1----- 1------1— 100
Figure 5.10
50
Simulated
0 Hz
-50
-100
NMR Spectra of a 2-Spin System as the Ratio
Av/J, is Varied from 10.0 to 0.0
63
Chapter 5 NMR Spectroscopy
C oin cid en ta lly e q u a l va lu es o f c o u p lin g constants.
An AMX spin system. First order analysis rules predict that the resonance for Hm in an AMX spin system will be a doublet of doublets (4 lines) since HMwill be split by coupling to Ha and to Hx- All lines of the multiplet will have equal intensity and the spacings in the multiplet will be J a m and J ax However, in the case where J am and J ax are approximately equal, the central lines of the multiplet overlap to give a single line whose intensity is twice as high as the outer lines.
While the multiplet is
A doublet of doublets appears like a triplet if
technically a doublet of
approximately equal
Jamand ^mxare
doublets, it appears as a triplet
Hm
(3 lines) with intensities in the ratio 1 :2 : 1 .
An A 2M 2X2 spin system. First order analysis rules predict that the resonance for Hm in an A 2M 2X2 spin system will be a triplet of triplets (9 lines) since Hm will be split by coupling to 2 x Ha nuclei and to 2 x Hx nuclei. The relative intensities of the lines in the multiplet can be predicted easily using Table 5.9. The spacings in the multiplet will be equal to J a m and J a x However, in the case where Jam and J ax are approximately equal, there is overlap between the lines of the multiplet. A triplet of triplets appears like a quintet if Jam and JMXare approximately equal
While the multiplet is technically a triplet of triplets, it appears as a
H*
quintet (5 lines) with 4I
intensities in the ratio 1:4:6:4 1.
1 2 Ll 1 2
1\
1 I
This is not an uncommon situation in flexible alkyl chains (X-CH2-CH2-CH 2-Y) since the 3 -bond vicinal coupling between protons on adjacent carbons typically falls within a narrow range of about 6-8 Hz.
64
Chapter 5 NMR Spectroscopy
(2) Spin Decoupling
In the signal of a proton that is a multiplet due to spin-spin coupling, it is possible to remove the splitting effects by irradiating the sample with an additional Rf source at the exact resonance frequency of the proton giving rise to the splitting. The additional radiofrequency causes rapid flipping of the irradiated nuclei and as a consequence nuclei coupled to them cannot sense them as being in either an a or p state for long enough to cause splitting. The irradiated nuclei are said to be decoupled from other nuclei in the spin system. Decoupling simplifies the appearance of complex multiplets by removing some of the splittings. In addition, decoupling is a powerful tool for assigning spectra because the skilled spectroscopist can use a series of decoupling experiments to sequentially identify which nuclei are coupled. In a 4-spin AM2X spin system, the signal for proton HAwould appear as a doublet of triplets (with the triplet splitting due to coupling to the 2 M protons and the doublet splitting due to coupling to the X proton). Irradiation at the frequency of Hx reduces the multiplicity of the A signal to a triplet (with the remaining splitting due to ./AM) and irradiation at the frequency of HMreduces the multiplicity of the A signal to a doublet (with the remaining splitting due to JAX) (Figure 5.11). Ha
Basic spectrum of Ha without irradiation of of H m or Hx
Figure 5.11
with irradiation of Hx
with irradiation of HM
Selective Decoupling in a Simple 4-Spin System
65
Chapter 5 NMR Spectroscopy
5.10
CORRELATION OF 1H - 1H COUPLING CONSTANTS WITH STRUCTURE
Interproton spin-spin coupling constants are of obvious value in obtaining structural data about a molecule, in particular information about the connectivity of structural elements and the relative disposition of various protons.
Non-aromatic Spin Systems. In saturated systems, the magnitude of the geminal coupling constant -VH_C_H(two protons attached to the same carbon atom) is typically between 10 and 16 Hz but values between 0 and 22 Hz have been recorded in some unusual structures.
FL
Ha C 'C
FT' n H b
2
Jab =10-16 Hz.
The vicinal coupling (protons on adjacent carbon atoms) V H.C.C.Hcan have values 0 - 16 Hz depending mainly on the dihedral angle for 90 < ()>< 180°
IN m8 < ^ 4 o
0
45 90 135 180 Dihedral angle(tj)) (degrees)
It follows from these equations that if the dihedral angle between two vicinal protons is near 90° then the coupling constant will be very small and conversely, if the dihedral angle Ai-c=c-H(tram) not overlap. This means that the stereochemistry of the double bond can be determined by measuring the coupling constant between vinylic protons. Where the C=C bond is in a ring, the 5J H.C=C.Hcoupling reflects the ring size. Ha
H \
J,'AB(cis)
/ Hb / C=C\ X Hc
%AC(trans) = 12-19 Hz J\BC(gem) = 0 - 3 Hz
= 5 - 7 Hz
JAB(cis)
= 6 - 11 Hz Hb Ha
I
=9-11
JAB(cis)
HZ
Hb
In alkyl-substituted alkenes, the long-range allylic couplings, (4J ab and 4J ac ) are typically in the range 0-3 Hz. In systems which are stereochemically constrained, the magnitude of the coupling is a function of the dihedral angle between the C-HAbond and the plane of the double bond in a
\
y ha C= C 7
\
He
4j AB] 4j AC = 0 - 3 Hz
relationship reminiscent of the Karplus relation.
67
Chapter 5 NMR Spectroscopy
Aromatic Spin Systems The coupling constant between protons attached to an aromatic ring is diagnostic of the relative position of the coupled protons i.e. whether they are ortho, meta or para. Ha
Ha
J,AB(meta)
= 1 - 3 Hz
AB(para)
0 -1 .5 Hz
Similarly in condensed polynuclear aromatic compounds and heterocyclic compounds, the magnitude of the coupling constants between protons in the aromatic rings reflects the relative position of the coupled protons.
SJ 1,2 = 8 .3 - 9 .1 Hz
3J2 ,3 = 4.0 - 5.7 Hz
3J2i3 = 4.7 Hz
3J2>3 = 1.8 Hz
3J2 ,3 = 6.1 - 6 . 9 Hz
3J3,4 = 6.8 - 9.1 Hz
3J3,4 = 3.4 Hz
3J3,4 = 3.5 Hz
4J1i3= 1 .2 - 1 .6 Hz
4J2 ,4 = 0.0 - 2.5 Hz
4J2,4 = 1.0 Hz
4J2,4 = 0.8 Hz
5J1i4 = 0 - 1 . 0 Hz
4J3,5 = 0 .5 - 1 . 8 Hz
4J2,5 = 2.9 Hz
4J2,5 = 1.6 Hz
5J1iS = 0 - 1 . 5 Hz
4J2,6 = 0.0 - 0.6 Hz
4J2,4; 4d2,6 = 0 Hz
d2i3 = 4.3 Hz
5J2 ,5 = 1.5 Hz
4J2 ,4 = 1.8 Hz
3J4 si 3J s,g = 5.0 Hz
3J3i4 = 8.3 Hz
5J2 ,5 = 0.0 - 2.3 Hz 4
n
h- n^
\
n
3
r\
M
5^
N H
4Jia H and 13C nuclei in a sample. The resulting 13C DEPT spectrum contains only signals arising from protonated carbons (non protonated carbons do not give signals in the 13C DEPT spectmm). The signals arising from carbons in CH 3 and CH groups (i.e. those with an odd number of attached protons) appear oppositely phased from those in CH 2 groups (i.e. those with an even number of attached protons) so signals from CH 3 and CH groups point upwards while signals from CH2 groups point downwards (Figure 6.1b). In more advanced applications, the 13C DEPT experiment can be used to separate the signals arising from carbons in CH3, CH2 and CH groups. This is termed spectral
72
Chapter 6 13C NMR Spectroscopy
editing and can be used to produce separate 13C sub-spectra of just the CH3 carbons, just the CH2 carbons or just the CH carbons. Figure 6.1 shows various 13C spectra of methyl cyclopropyl ketone. The 13C spectrum acquired with full proton decoupling (Figure 6.1a) shows 4 singlet peaks, one for each of the 4 different carbon environments in the molecule. The DEPT spectrum (Figure 6.1b) shows only the 3 resonances for the protonated carbons. The carbon atoms that have an odd number of attached hydrogens (CH and CH3 groups) point upwards and those with an even number of attached hydrogen atoms (the signals of CH2 groups) point downwards. Note that the carbonyl carbon does not appear in the DEPT spectrum since it has no attached protons. In the carbon spectrum with no proton decoupling (Figure 6.1c), all of the resonances of protonated carbons appear as multiplets and the multiplet structure is due to coupling to the attached protons. The CH3 (methyl) group appears as a quartet, the CH2 (methylene) groups appear as a triplet and the CH (methine) group appears as a doublet while the carbonyl carbon (with no attached protons) appears as a singlet. In Figure 6.1c, all of the !JC_Hcoupling constants could be measured directly from the spectrum. CH, (c) with 1H fully coupled
o II c
CH CH c h 2^
;c=o
- ch /
x ch3
ch2
(b)
DEPT
ch2(,
CH3f CHf
(a) with 1H fully decoupled
I 215
205
Figure 6.1
30
T— i 20
10
r ppm
13C NMR Spectra of Methyl Cyclopropyl Ketone (CDCI3 Solvent, 100 MHz), (a) with Broad Band Decoupling of *H ; (b) DEPT Spectrum (c) with no Decoupling of 1H.
73
Chapter 6 13C NMR Spectroscopy
For purposes of assigning a 13C spectrum, two 13C spectra are usually obtained. Firstly, a spectrum with complete *H decoupling to maximise the intensity of signals and provide sharp singlets to minimise any signal overlap. This is the best spectrum to count the number of resonances and accurately determine their chemical shifts. Secondly, a spectrum which is sensitive to the number of protons attached to each C to permit partial sorting of the 13C signals according to whether they are methyl, methylene, methine or quaternary carbon atoms. This could be a DEPT spectrum or a 13C spectrum with no proton decoupling. The number of resonances visible in a 13C NMR spectrum immediately indicates the number of distinct 13C environments in the molecule (Table 6.1). If the number of ,3C environments is less than the number of carbons in the molecule, then the molecule must have some symmetry that dictates that some 13C nuclei are in identical environments. This is particularly useful in establishing the substitution pattern (position where substituents are attached) in aromatic compounds.
Table 6.1
The Number of Aromatic 13C Resonances in Benzenes with Different Substitution Patterns
Molecule
o o-° a:
Number of aromatic ,3C resonances
1
Cl—
4
Br^
\
74
Cl
Number of aromatic 13C resonances
—Cl
2
a
4
y
Cl 3
/Cl
0
Molecule
0
6
Br Cl
4
Br
6
Chapter 6 13C NMR Spectroscopy
6.3
SHIELDING AND CHARACTERISTIC CHEMICAL SHIFTS IN 13C NMR SPECTRA
The general trends of l3C chemical shifts somewhat parallel those in 'H NMR spectra. However, l3C nuclei have access to a greater variety of hybridisation states (bonding geometries and electron distributions) than 'H nuclei and both hybridisation and changes in electron density have a significantly larger effect on l3C nuclei than 'H nuclei. As a consequence, the l3C chemical shift scale spans some 250 ppm, cf. the 10 ppm range commonly encountered for 'H chemical shifts (Tables 6.2 - 6.4, Figure 6.2). Table 6.2
Typical l3C Chemical Shift Values in Selected Organic Compounds Compound
8 ,3C (ppm from TMS)
CH4 CH3CH3 CH3OH CH 3CI CH2C12 CHCl,
-2.1 7.3 50.2 25.6 52.9 77.3
CH 3CH2CH2C1
11.5 (CH3) 26.5 (-CH2-) 46.7 (-CH 2-C1)
c h 2= c h 2
123.3
c h 2= c = c h 2
208.5 (=C=) 73.9 (=CH2)
CH3CHO
31.2 (-CH3) 200.5 (-CHO)
CH 3COOH
20.6 (-CH3), 178.1 (-COOH)
CH3COCH3
30.7 (-CH3), 206.7 (-CO-) 128.5
0 3 4 0\ = /
2
149.8 (C-2) 123.7 (C-3) 135.9 (C4)
75
Chapter 6 13C NMR Spectroscopy
Table 6.3
Typical ,3C Chemical Shift Ranges in Organic Compounds
Group
13C shift (ppm)
TMS
0.0
-CH 3 (with only -H or -R at Ca and Cp)
0-30
-CH2 (with only -H or -R at Ca and Cp)
20-45
-CH (with only -H or -R at Ca and Cp)
30-60
C quaternary (with only -H or -R at Ca and Cp)
30-50
o -c h 3
50-60
n -ch 3
15-45
oc
70 - 95
c=c
105 - 160
C (aromatic)
110- 155
C (heteroaromatic)
105 - 165
-ON
115 - 125
C=0 (acids, acyl halides, esters, amides)
155 - 185
C=0 (aldehydes, ketones)
185 -225
In 13C NMR spectroscopy the l3C signal due to the carbon in CDC13 appears as a triplet centred at 5 77.0 with peak intensities in the ratio 1:1:1 (due to spin-spin coupling between 13C and 2H). This resonance serves as a convenient reference for the chemical shifts of 13C NMR spectra recorded in this solvent. Table 6.4 gives characteristic 13C chemical shifts for some s/Ahybridised carbon atoms in common functional groups. Table 6.5 gives characteristic 13C chemical shifts for some .syr-hybridised carbon atoms in substituted alkenes and Table 6.6 gives characteristic l3C chemical shifts for some .s/j-hybridised carbon atoms in alkynes.
240
Figure 6.2
220
200
180
160
140
120
100
80
60
Approximate 13 C Chemical Shift Ranges for Carbon Atoms in Organic Compounds
40
20
0
-20
5 (ppm)
Chapter 6 13C NMR Spectroscopy
77
Chapter 6 13C NMR Spectroscopy
CH3 CH2-
c h 3— X
AA
13C Chemical Shifts (5) for sp3 Carbons in Alkyl Derivatives (CH3 )2 CH-— X
— ch3
— ch3
— H
-2.3
7.3
7.3
15.4
15.9
— ch=ch2
18.7
13.4
27.4
22.1
32.3
— Ph
21.4
15.8
29.1
24.0
34.3
— Cl
25.6
18.9
39.9
27.3
53.7
— OH
50.2
18.2
57.8
25.3
64.0
— och3
60.9
14.7
67.7
21.4
72.6
— OCO-CH 3
51.5
14.4
60.4
21.9
67.5
30.7
7.0
35.2
18.2
41.6
20.6
9.2
27.2
19.1
34.1
— nh2
28.3
19.0
36.9
26.5
43.0
— NH-COCH3
26.1
14.6
34.1
22.3
40.5
— CEN
1.7
10.6
10.8
19.9
19.8
— no2
61.2
12.3
70.8
20.8
78.8
CO
X
X
0 1 0 0
X i ro 1
Table 6.4
— CO-OCH 3
Table 6.5
— ch3
13C Chemical Shifts (8 ) for sp2 Carbons in Vinyl Derivatives: CH 2=CH-X X
1 X
II
115.9
136.2
- C ( C H 3 )3
108.9
149.8
— Ph
112.3
135.8
116.3
136.9
129.2
117.3
128.0
137.1
— CO-OCH 3
130.3
129.6
— Cl
117.2
126.1
—o c h 3
84.4
152.7
—0 C0 -CH3
96.6
141.7
— CEN
137.5
108.2
—n o 2
122.4
145.6
— N(CH3)2
91.3
151.3
CM
II
X
— ch3
0
123.3
X
123.3
0
—H
1
X
0
III
1
0
CO
X
0 1 0 0
78
0
c h 2=
X
>
H~
Chapter 6 13C NMR Spectroscopy
Table
13C Chemical Shifts (8 ) for sp Carbons in Alkynes:
6 .6
X-C=C-Y
X —c = H—
—H
H—
—
66.9
79.2
67.0
92.3
80.0
82.8
66.3
67.3
77.1
83.4
81.8
78.1
o c h 2c h 3
22.0
88.2
ch3
72.6
72.6
Ph
79.7
85.8
97.4
87.0
89.4
89.4
74.6
74.6
ch3
c h = ch2
H—
1
—
I
III
H—
0
0
H—
Ph
—coch3 —
c h 3—
—
c h 3—
I
0
73.2
- C ( C H 3)3
H—
H—
73.2
—
H—
—
1CO
—coch3
Ph — CO O CH3
—
—
= C -Y
Ph
cooch
3
Table 6.7 gives characteristic 13C chemical shifts for the aromatic carbons in benzene derivatives. To a first approximation, the shifts induced by substituents are additive. So, for example, an aromatic carbon which has a -N O 2 group in the para position and a -B r group in the ortho position will appear at approximately 137.9 ppm [(128.5 + 6.1(p-N02) + 3.3(o-Br))].
79
Chapter 6 13C NMR Spectroscopy
Table 6.7
Approximate 13C Chemical Shifts (8 ) for Aromatic Carbons in Benzene Derivatives Ph-X in ppm relative to Benzene at 128.5 ppm (a positive sign denotes a downfield shift)
8
X
ipso
ortho
meta
para
— H
0.0
0.0
0.0
0.0
— no2
19.9
-4.9
0.9
6.1
2.0
1.2
-0 . 1
4.3
5.0
- 1. 2
0.1
3.4
8.9
0.1
- 0. 1
4.4
— CEN
-16.0
3.5
0.7
4.3
— Br
-5.4
3.3
2.2
- 1.0
8.9
-2.3
-0.1
-0 . 8
— Cl
5.3
0.4
1.4
-1.9
— ch3
9.2
0.7
-0. 1
-3.0
— o co -C H s
22.4
-7.1
0.4
-3.2
— och3
33.5
-14.4
1.0
-7.7
— nh 2
18.2
-13.4
0.8
- 1 0 . 0
-C O -O C H 3 CNJ
z
X
1
1
0 0
CO
X
O
1
O O
CNJ
X
O
II X
0
Table 6.8 gives characteristic shifts for 13C nuclei in some polynuclear aromatic compounds and heteroaromatic compounds. Table
6 .8
Characteristic 13C Chemical Shifts (8 ) in some Polynuclear Aromatic Compounds and Heteroaromatic Compounds
80
7 2-Dimensional NMR Spectroscopy A two-dimensional NMR spectrum has two frequency axes rather than one. A 2D spectrum is acquired using a pulse sequence which contains a delay period 'tf which can be varied systematically as the experiment is repeated. The experiment is repeated many times (typically 512 or 1024), with a different delay V in the pulse sequence for each experiment. One FID is acquired for each experiment giving an array of'N ' individual FID's each of which has been acquired with a slightly different pulse sequence. Each FID represents the variation of detected signal as a function of time (t2 in the diagram below) and successive FIDs in the array differ as a function of the time variable ti within the preparation period of the pulse sequence. Fourier transformation of the two-dimensional array of data with respect to t2 affords a series of spectra which vary systematically as a function o f f . A second Fourier transformation, this time with respect to ti, gives a two-dimensional spectral array (which is function of two frequency domains F t and F2). Two-dimensional spectra are usually represented in terms of a stacked plot or contour plot. The contour plot is a more convenient representation for making measurements or peak assignment.
Time dependent pulse sequence
time t2
frequency F2
FID
frequency F2
Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
81
Chapter 7 2-Dimensional NMR Spectroscopy
The number of possible two-dimensional experiments is essentially unlimited. Different pulse sequences in the preparation period give rise to different twodimensional spectra which can be tailored to exhibit various properties of the sample. The technical detail behind multi-dimensional NMR experiments and the pulse sequences used to generate 2D spectra, is beyond the scope of this book.
Two-dimensional spectra have the appearance of surfaces, generally with two axes corresponding to chemical shift and the third (vertical) axis corresponding to signal intensity. s h ift ( f T)
It is usually more useful to plot twodimensional spectra viewed directly from above (a contour plot of the surface) in order to make measurements and assignments.
CF®
to
ra
o E Q) . O
Chemical shift (F2 )
The most important two-dimensional NMR experiments for solving structural problems are COSY (C o rrelation SpectroscopY), NOESY (Nuclear Overhauser Enhancement SpectroscopY), HSQC (Heteronuclear Single Quantum Correlation) or HSC (Heteronuclear Shift Correlation), HMBC (Heteronuclear Multiple Bond Correlation) and TOCSY (TOtal Correlation SpectroscopY). Most modem high-field NMR spectrometers have the capability to routinely and automatically acquire COSY, NOESY, HSQC, HMBC and TOCSY spectra.
82
Chapter 7 2-Dimensional NMR Spectroscopy
7.1
COSY (CORRELATION SPECTROSCOPY)
The COSY spectrum shows which pairs of protons in a molecule are coupled to each other. The COSY spectrum is a symmetrical spectrum that has the 'H NMR spectrum of the substance as both of the chemical shift axes (Fi and F2). It is usual to plot a normal (one-dimensional) NMR spectrum along each of the F 1 and F2 axes to give reference spectra for the peaks that appear in the two-dimensional spectrum. A COSY spectrum of 1-iodobutane (CH 3CH 2CH2CH2I) is given below (Figure 7.1). F2 Axis
1H NMR Spectrum 4
J
J ____ k
O (D Q_
CO
(Z
(J) x
<
Figure 7.1
*H COSY Spectrum of 1-Iodobutane (CDCI3 solvent, 298K, 600 MHz)
The COSY spectrum of 1-iodobutane has a set of 4 peaks on the diagonal as well as peaks that are off the diagonal. The COSY spectrum is always symmetrical about the diagonal - the off-diagonal peaks above the diagonal are mirrored on the lower side of the diagonal. The off-diagonal peaks are the important signals, since these occur at positions where there is coupling between a proton on the F 1 axis and a proton on the
83
Chapter 7 2-Dimensional NMR Spectroscopy
F2 axis. The protons which are part of the -CH2I group (Hi) are easy to identify since the halogen substituent characteristically moves these to about 8 3.2 ppm. There are 3 off-diagonal peaks on each side of the diagonal - one indicates the coupling between Hi and H2; the second indicates coupling between H2 and H3 and the third indicates coupling between H3 and H4. If you can identify one of the proton signals, then you can identify the protons that are coupled to it and then work sequentially along a coupled network until all the protons which are coupled together are identified. Hi - * - H 2
I
H3—*-H 4
H I1
H I2
H H I3 U
H
H
H H
C— C— C— C—H O ' h 2- *
- h3
In a single COSY spectrum all of the coupling pathways in a molecule can be identified.
7.2
THE HSQC (HETERONUCLEAR SINGLE QUANTUM CORRELATION) OR HSC (HETERONUCLEAR SHIFT CORRELATION) SPECTRUM
The HSQC spectrum or the HSC spectrum are the heteronuclear analogues of the COSY spectrum and these experiments identify which protons are directly attached to which carbons in the molecule. The HSQC spectrum has the 'H NMR spectrum on one axis (F2) and the BC spectrum (or the spectrum of some other nucleus) on the second axis (Fi). It is usual to plot a normal (one-dimensional) 'H NMR spectrum along the proton dimension (the F2 axis) and a normal (one-dimensional) 13C NMR spectrum along the 13C dimension (the Fi axis) to give reference spectra for the peaks that appear in the two-dimensional spectrum. An HSQC spectrum of 1-iodobutane (CH3CH2CH2CH 2I) is given in Figure 7.2. The HSQC spectrum does not have diagonal peaks. The peaks in an HSQC spectrum occur at positions where a proton in the spectrum on the F2 axis is directly coupled to a carbon in the spectrum on the Fi axis via a 1-bond C-H coupline.
84
Chapter 7 2-Dimensional NMR Spectroscopy
1H NMR Spectrum
F2 Axis
4
J
2
3
k
A
ppm
E
5
o
< D Q_ CO
10
C£ z:
o
15
CO
20
25
30
c/) x <
Figure 7.2
35
*H - 13C HSQC Spectrum of 1 -Iodobutane (CDCI3 solvent, 298K, ’H 600 MHz, 13C 150 MHz)
1-Iodobutane has 4 carbon resonances and the h3- * - c 3
HSQC spectrum shows 4 cross peaks. Having identified all of the resonances in the !H spectrum, then the resonances in the carbon
Ch
H *W I2 V I 3
H
U C— C— c — H
1C 1
spectrum can simply be identified by the
H
positions of the cross peaks corresponding to
7
H4—*-C 4
each proton resonance, In the HSQC or HSC experiment, all of the protons which are coupled to carbons can be identified. It is usually possible to assign all of the resonances in the ]H NMR spectrum i.e. establish which proton in a molecule gives rise to each signal in the spectrum, using spin-spin coupling information or a COSY experiment, then assign the signals in the l3C spectrum by correlation to the known proton resonances. Note that non-protonated carbons (quaternary carbons) do not give rise to signals in the HSQC or HSC spectra.
85
Chapter 7 2-Dimensional NMR Spectroscopy
7.3
HMBC (HETERONUCLEAR MULTIPLE BOND CORRELATION)
The HMBC spectrum correlates chemical shifts of hydrogen nuclei with carbon nuclei which are separated by two or more chemical bonds. The HMBC experiment is frequently used to assign quaternary and carbonyl carbons which don’t have any directly bound protons so they are “invisible” in the HSQC or HSC experiment. The HMBC experiment is designed to filter out correlations resulting from large C-H coupling constants (120-160 Hz) resulting from protons directly bound to carbon and to select for smaller couplings (around 10 Hz) which are typical C-H couplings over 2 or 3 bonds. The HMBC is a very powerful method for making the connection between two parts of a molecule that may be isolated from each other by a carbonyl, an ether, an ester, an amide or by some other functional group. ^
X
H
\H
______ c — x— ,H X = 0, S,NH
1, It is usual to plot a normal (one-dimensional) H NMR spectrum along the proton
dimension (the F2 axis) and a normal (one-dimensional) 13C NMR spectrum along the 13
C dimension (the F 1 axis) to give reference spectra for the peaks that appear in the
two-dimensional spectrum. A HMBC spectrum of 1-iodobutane (CH3CH2CH2CH2I) is given in Figure 7.3. The HMBC spectrum does not have diagonal peaks. The peaks in an HMBC spectrum occur at positions where a proton in the spectrum on the F2 axis is coupled to a carbon in the spectrum on the Fi axis via a 2-bond or 3-bond CH coupling.
86
Chapter 7 2-Dimensional NMR Spectroscopy
1H NMR Spectrum
F2 Axis
A
J
JL
L
£ 3— L o 0
Q-
(/) q: z
o co
-
CO CD CL (D (/) CO _Q 7 0 0^
45
. 40 — 20
No significant UV
M+’
absorption above 220 nm
74 57
li
k
it
1
1 ___ L-
c 3 h60
2
_J— 1— 1— i— 1— 1— 1__1__1__1— 1__1— 1— 1— 1__1— 1__1__1__1__1__1__1__1__
80
40
120
160
200
240
280
m/e 1
1
■
1
1
1—
f
1
1
•
1
1
1
1
1
1
1
.
1
1
1
1
1
■
1
•
1
•
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
c h 2|
CH3f a ■4
solvent
proton decoupled
hi
.............................................................................
200
.
160
I
.
1
.
120
I
80
40
1 -
0
— 5 (ppm)
• ... -
1H NMR Spectrum (200 MHz, CDCI3 solution) exchanges with D20
_A_
12
11
i
i 10
9
8
7
6
5
4
3
2
1
TMS
0 6 (ppm)
113
Problem 3
4000
3000
2000 1600 V (cm'1)
1200
800
0.0
UV spectrum solvent: ethanol 15.4 mg /1 0 mis path length : 1.00 cm
40
120
80
160
240
200
280
250
200
T
' 1
"1..—
I
'
I
'
350
300 X (nm)
m/e I
'
i
■
1
■
1
'
I
'
1
1
1
i
1
1
1
1
1—
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT CH2| CH3| CHf
proton decoupled
solvent III
1 1
1
1
i
1
200
1H NMR
1
1
160
1
1
1
1
120
1
1
1
1
80
1
1
40
1
0
5 (ppm)
Spectrum
(200 MHz, CDCIj solution)
r 1 J
10
114
8
i
L
0 6 (ppm)
Problem 4
4000
3000
2000
1600
1200
800
V (cm'1) Mass Spectrum
57
29 ■ 0 -. Q 03 3. ~ Q 3 CO -. _0Q 3 7O - sS
No significant UV
00 0 0
II +
2
absorption above 220 nm
C 4H 8O2 L ii i » 1 j—i_i_J ____1— 1___ *-- 1-- *-- »-- *-- 1-- 1-- »-- •-- 1-- •-- 1-- *-- •-- •-- 1-- •-- 1—- i • * *
40 1
80 1
'
120
' ... 1
1
160 m /e
1
' 1
200 '
240
1
'" " " T
280 1
1
'
,
1
.
|
1
|
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
1
..................
DEPT
CH2(| CH3f C H f
proton decoupled
solvent I
I
■
I
■
I
200
10
9
■
I
.
I
160
8
7
■
1
■
1 ..... X
120
6
5
1
■
1 — I------- \-------1------ 1------- 1-------1-------1----------------
80
4
40
3
2
0
1
S(PPm)
0 6 (ppm)
115
Problem 5
1600 2000 V (cm'1)
3000
4000 100 :
Mass Spectrum
27
80 60
800
1200
107/109
- ^ • CO - 0 . Q_ “ 0
No significant UV
1 0 . _Q
40 r
absorption above 220 nm
o
M+ ’
- osO N
186/188/190
20
,i, C 2 H 4 B r2 1 U, At. .1 i, ■ » *__I__.__I_____I___—I__1—«— 1—I— 1—I— 1—I—*—i—•— 1— 1— 1—>— 1—■— 1—•— 1—'—
120
80
40
160
240
200
280
m/e ------------- I—
'—
i—
i
>-
1
i
>
i
1
i
i
■
i
■
i
■
i
1
. 1
'
1
'
.
1
.
"1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
solvent |
proton coupled
proton decoupled
I
.
I
^
*
1
1
1
200 1H NMR
1
1
160
-1
J ___ j
1
1
120
L
80
40
I
0
5 (ppm)
Spectrum
(200 MHz, CDCI3 solution)
TMS
...............
j
I
10
116
1
I
9
I
»
8
I
1
7
1
- t ------ 1-------- 1-------- 1 ----- -—1-------- 1-------- •-------- 1 -------- •-------- 1-------- 1-------- 1 -------- 1-------- 1 -----
6
5
4
3
2
1
0 8 (ppm)
Problem 6
2000
3000
4000
1200
1600
800
V (cm'1)
100
Mass Spectrum
43
80 -■ CO -. Q CD. D 60 -“ cC n - C O . _Q 40 7 o
UV Spectrum ^max 289 nm (log^gS 1.4) solvent: methanol
C CO O II + 2
20 r i. i C 4 H6 ° 2 __1___1___L » i_u—1_1—1_1_1_1_1_1_1__ _l-- .-- 1-- ,-- 1-- .-- 1-- .-- 1-- .—J- —■—i 280 200 240 120 160 40 80
m /e 1
’
1
1
1
1
1
1
’
1
1
1
|
1
1
1
.. 1
r
1. i
1
13C NMR Spectrum (100 MHz, CDCIj solution)
D EP T
c h 2| c h 3| c h |
proton decoupled
solvent
* I
.... i
i
i
i
i
i
200
i
i
i
i
i
160
i
i
i
i
i
120
.................. i
i
.j
_i-------1-------1-------1-------1-------1-------1----------------
80
40
0
5 (ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
TMS
— I------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1___
10
9
8
7
6
5
1
-
4
•
1
3
1
«
2
1
.
1
1
1
1
0 8 (ppm)
117
Problem 7
4000
3000
2000 1600 V (cm'1)
800
1200
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e --------------i—
i—
\—
i—
r
1
' ...i
1
i
i
r '"
1
i
1
i
1......... i
i
■
i
■
.
I
,
i
1
I
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
c h 2| CH3f
cuf
proton decoupled
I
,
I
I I
.
I
.
I
200
.
I
.
160
solvent 1
I
.
I
.
120
I
.
I
.
I
80
40
0
S(ppm)
I
i
1H NMR Spectrum (400 MHz, CDCI3 solution)
I
10
118
i
I
9
i
I
8
'
I
7
'
I
6
i
5
I
i
4
I
i
3
!
i
2
I
i
1
L
0 6 (ppm)
Problem 8
4000
3000
1600
2000
800
1200
V (crrf1 )
No significant UV absorption above 220 nm
80
40 1 1 , . 13C NMR Spectrum
120 1
,
160 m /e
1
1
1
240
200
,
j
1
280
1
1
1
1
1
1
1
1
■
1
1
( 5 0 .0 M H z , C D C I 3 so lu tio n )
DEPT
c h
2|
C H 3f
CHf
p ro to n d e c o u p le d
I
.
s o lv e n t
I . 200
i---- 1
i i i
i
160
i i----1 120
i
III
I
i . 80
i--- 1---- 1---- 1---- 1---- 1---- 1 _i-------------------40 0 5 (ppm)
1H NMR Spectrum (200
M H z,
CDCIg s o lu tio n )
TM S
i
i
1
1
1
1
1
------ ------------ ------------ ------------1------------ ------------1----------- ----------- 1—
10
9
8
7
6
i
«
5
1
•
4
i
1
3
1
i
2
i
1
1
1
i
1
0 5 (ppm)
119
Problem 9
100 :
42
Mass Spectrum
80 “• CO - CD . Q. CD _ CO . _Q
No significant UV
60 -
40 7
M+55
O
absorption above 220 nm
70
- -5
20 C 5 H 10
: I, 1 ___ ._1_1_L _J-- 1.
40
.
.
1
.
.
1
1
120
80
i-- .-- 1-- .-- 1-- .---- 1-- .---1-- .-- 1-- .-- 1-- .-- 1-- .--
.
200
160
240
280
m/e 1.......' MMR (1 2 5 .0
MHz,
1 ..T ■"" I
1
1
1
1
' .....1
.9 n p r .tr n m
1
1
' ' .... ”T.......r.... 1
1
1
1
1
1
1
'
I l 40
I
1
1 0
l 8 (ppm)
_
C D C I 3 solutio n)
DEPT
CH2{ CH3f C H f
pro to n d e c o u p le d
s o lven t
1
l
I
l
I l 200
1
l
1 l 160
I
1
I l 120
I
1
I i 80
I
l
1H NMR Spectrum ( 4 0 0 M H z , C D C I 3 solutio n)
_J
10
120
L_
-I
I
8
i
L
7
_i
L
1
0 8 (ppm)
Problem 10
4000
1600 V (cm‘1)
3000
800
1200
2000
100 :
Mass Spectrum
59
80 " ro . C0
l
No significant UV
60 * 00 -
40
. _CQO 7o - £5
absorption above 220 nm M +’=118 (
8
1
.
7
1
.
6
1
5
1
.
4
1
.
3
1
.
2
1
i
1
1
1
1
0 5 (ppm)
Problem 54
4000
2000
3000
1200
1600
800
V (cm'1) 100
152
-
Mass Spectrum
“ •
CD in
:
40 7 -
20
76
-0 0
00
CO CD
CM CD
60
■ -
X CM X—
0
80
UV Spectrum
M
A,max 264 nm (log10e 4.5)
+ '
solvent: methanol
310/312/314
•
i | i
-t 1 1
1 .II, Jlil .il 1, l ------ .-- 1-- .---1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- 1-- .-- L--. 1 . 1 . 1 . 40 80 120 160 200 240 280
m/e I
I
'
I
'
' 1
1
«
I
|
.
j
,
1 ■ ,
|
.
j
1
|
,
1
.
1
.
|
.
1
.
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2)|f CH3f
ch
|
solvent
proton decoupled
i
•
I
I
I
1
200
-1
1 .. 1
160
1
1
1
.
.
120
1
1
.
1
80
1H NMR Spectrum
40
0
8
(ppm)
expansion
(400 MHz, CDCI3 solution)
residual solvent
7.8
7.6
7.4
7.2
ppm
TMS
-J----- 1----- 1___ I___ I___ 1
- J _________ I_________ I_________ I_________ L_
10
9
8
7
6
5
4
1 3
_J_________ I_________ I_________ I_________ L_
2
1
0 8 (ppm)
165
Problem 55
4000
3000
2000 1600 V (cm'1 )
100 :
1200
800
Mass Spectrum
125/127
8 0 -• CO -
CD
6 0 -■ 5
-. CL CP CP 60 “ ■ C/3 -. -O CO 40 7 o - -s
20 C 12H 6 1 il II il Jl l III J ------ 1---1—1—1—1—1—1_1_1_1_l_l_1_1_■ —i—i—i—i_i_i—i_i_i_i_i_i_i_■ i ■
120
80
40
200
160
240
280
m/e i
i
1
I
•
1
I
1
I -1
1
|
|
.
|
1 .... 1
1
1
1
1
1
1
1
1
*
1
1____I____1_________
13C NMR Spectrum (75.0 MHz, CDCI3 solution)
D EPT
CH2J CH3f C H f
1—v- solvent
proton decoupled
1
l
*
.
I
1
.
200
•
l
160
*
l
1
.
.
120
I
1
•
«
80
40
0
5 (ppm)
1H NMR Spectrum (300 MHz, CDCI3 solution)
f.......
solvent residual
TMS
r 1 .
i
1
10
.
1
9
1
.
8
1
1------- 1------- 1------- 1------- 1—
7
6
5
1
1
4
1
1
3
1
1
2
1
1
1
......1_____..........1 .
0 5 (ppm)
171
A-(nm)
m/ e — .. . r
'
1
i
i
1
i
1
i
^
1
1
1
1
1
'
1
1 .. 1
1
1
1
1
1
1
1
1
1
1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
ch 2| ch 3| c h |
solvent
proton decoupled
1 ----- 1------1----- »----- L----- 1------1----- 1----- 1----- 1------1------1------1----- 1_— 1-----j
200
160
120
1
1
1
80
40
0
8
(ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
___4 TMS L _ I
10
172
1
1
9
1
1
8
—1------- 1------- 1------- 1------- 1 — 1----- 1
7
6
5
4
1
1
3
1
1
2
_
1
1
1
1
1
1
0 8 (ppm)
Problem 62
4000
1200
2000
1600 V (cm'1 )
3000
100-
800
Mass Spectrum
105
UV Spectrum
80 Xmax 261 nm (log10s 2.7) 60 ^max 269 nm (log^s 2.5 )
40
M+,= 120 solvent: methanol
20 C9H12 40
80
120
160
200
240
280
m/e
1---- 1----1----1----1
■ ---- 1----1---- 1----1---- 1----1---- 1----'---- 1----•---- 1---- •---- 1----1---- 1----■ ---- 1----r
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EPT
CH2| CH3f
ch
|
i
solvent
proton decoupled
I
i
10
■ i . 200
9
i
8
■ i . 160
7
i
6
....................
.
i ■ i 120
5
..........1 ................
,
4
i 80
.
3
i
_
■ i 40
2
....... _
■ i
1
1
.
i 0
5
(ppm)
0 8 (ppm)
173
Problem 63
100
105
Mass Spectrum
UV Spectrum
80 A-max 270 nm (log10£ 2 .6 )
60 - 0CO
M + ' = 120
CO
solvent: methanol
40
20 C' 9q n H,12 _J
I
I
I
I
I
I
I
I
40
I
I
i
I
I
I
I
120
80
L_
-I
200
160
■
I
i
I
i
240
I
i
L_
280
m/e 13C NMR Spectrum MHz, CDCIj solution)
expansion
(100.0
11I' ‘1’I11' 1I11 135 130
DEPT
ch 2|
aJII. ' i 1i ' i i ■i ■
PPm
24
CH3f CHf expansion
135 130
JU i " i ■-11i 1i 1
solvent
ppm
24
20
J
200
ppm
expansion
11I■’1’T'... I
proton decoupled
20
160
120
ppm
I
I
80
I
I
1
40
I
0
L.
5 (ppm)
1H NMR Spectrum (400
I 10
174
MHz, CDCI3 solution)
■
I 9
i
I 8
■
I 7
i
I 6
I 5
I
I 4
I
I 3
I
i 2
I
i 1
L 0 5 (ppm)
m /e
I
1 l
1
l
'
l
1
X (n m )
l
1
l
1
I
1
i
i
1
i
1
i
1
1
r
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EP T
CH2| CH3f C H f
solvent proton decoupled
_JL_
.............................. I__ I__ I __ I__ I__ I__ I__ I__ 200 160 120
I ' l l __ I__ I__ I__ 1__ I__ L. 80 40 0 8 (ppm)
1H N M R S p e c tru m (400 MHz, CDCI3 solution)
r
f TMS L ----- 1--------- 1--------- 1--------- 1_____ t_____ L
10
9
8
1
7
1
i 1
1
6
.
5
1
.
4
1
.
3
1
1
2
1
,
1
1
,
1
0 8 (ppm)
175
Problem 65
IR Spectrum
802
(liquid film)
_L
_L
4000
3000
2000
1200
1600
800
V (cm1)
120
80
40
200
160
240
280
m/e T
' '
1.. “ I
1“
.... 1
'
' ~ 1— r_- i
1
r....•
1
1
•
1
•
1
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
CHa| CH3| O l f
solvent proton decoupled
- ___ I—.. 1
I
1 ......*
l
i
l
.
l
________________J L ______ ___ ________
1
t
160
200
1
.
80
120
1
........................................................
40
0
8
(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
r 1
"
TMS
-> u
J 10
176
i
I 9
i
I 8
i
I 7
i
I 6
i 5
I
■ 4
I
i 3
1
—. V
...
I
i 2
I
i 1
L 0 6 (ppm)
Problem 66
2000
1600 V (crn1 )
3000
4000
100
119
1200
800
Mass Spectrum
UV Spectrum
80 -
. CD Q_
^max 274 nm (log10s 2.3)
M+* = 134
60 - 0
^max 277 nm (log^s 2.3)
40
solvent: cyclohexane
20 C 10H 14
-J ■ --1--1--1--.--1-- I--1--1_ I_._I___I_............... 40
80
120
160
200
240
280
m/e *C NMR Spectrum (100.0 MHz, CDCI3 solution) expansion
DEPT
130
140
ppm 20
CH2)|f CH3f C H f
expansion
l
■ I 200
ppm
expansion
_ k jL proton decoupled
15
solvent
140
130
I
I 160
ppm
20
120
80
15
ppm
40
0
5 (ppm)
5 (ppm)
177
Problem 67
UV Spectrum max
278 nm
2.4)
X max 274 nm (log10s 2.4) A,max 270 nm (log10s 2.5) solvent: methanol
120
80
40
200
160
240
280
m/ e ----------- r ..1 i ' 13C NMR Spectrum
1
i
i'
■
i
•
1
i
1
i
,
r
|
1 "7
1
, ___1
'
1
1
1
1
,
1
1
(50.0 MHz, CDCI3 solution)
proton coupled
I
sol'vent
j
proton decoupled
I
.
I
1
.
.
I
200
.
I
160
i - .... 1..... 1
l
i
l
120
,
80
40
0
S(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
TMS .
I
10
178
1
I
9
1
I
8
1
I
7
J--------- 1--------- 1----
1
6
5
l
J
J--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1-----
4
3
2
1
0 8 (ppm)
Problem 68
UV Spectrum
X max 232 nm (log10s 3.4) ^max 248 nm (log^E) 3.5)
^max 265 nm (log^s 3.4) solvent: isooctane
40
120
80
160
200
240
280
m/e 1
1
1
1
T
1
1
'
1
'
1
'
1
'
7 ... t
1
1------- I
13C NMR Spectrum
1
1
'
1
1
expans ion
( 100.0 MHz, C D C I3 solution)
J
1.
1 1 1 1.... 1 D EP T
15
C H 2j CH3f c h |
13
ppm
expans ion
solvent i
proton decoupled
15
___________
1
.
1
200
.
1
.
1
160
1
_________
1
...
.L
T--- T---T,,,1---- 1
i
«
l
120
i
l
1
.
1
80
___ .... 1
1
13 1
40
ppm 1
1___.___ 1___ 1_______
0
s (ppm)
8 (ppm)
179
Problem 69
No significant UV absorption above 210 nm
40
120
80
200
160
240
280
m/e "I— 1----1— '— T 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
proton coupled
X solvent proton decoupled
i
10
180
120
160
200
8
6
80
40
0
1
5 (ppm)
0 8 (ppm)
Problem 70
4000
3000
2000
1200
1600
800
V (cm'1)
No significant UV absorption above 220 nm
40
120
80
160
200
240
280
m/ e 1
1
I
1
1
1
'
r"
1....... 1
' 1
1
1
1
'
'
"
1
'
1
..'
1
'
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
w
D EP T
nr, -r™,
C H C H 3f CHf
solvent proton decoupled
.
- - ....... -....-...... .. « 1 1 1
1
I ____ _____............ _......... ........... _ 1----- •- 1----- 1 1 1 1 1
200
160
...... I L _
1
120
1
1
1
1----- 1----- 1----- 1------ 1--
80
40
1 - 1
0
6
(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
exchanges with D20
TMS
J 10
i
I 9
i
I 8
,
L 7
-I_____ I_____ I_____ I_____ l_
6
5
4
J
i 3
I
i 2
I
i 1
L 0 6 (ppm)
181
Problem 71
4000
3000
2000
1600
1200
800
V (cm'1)
120
240
160
280
m/e i ..
'
i
1
r " ... 1
r "
X (nm) t
1
1
1
1
'
1
1
1
1
1
1
1
• .....r
.
1
.
1
,
1
,
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
ch
2| C H 3f c h |
proton decoupled
solvent
1 -1
-L -
*
J___ j
200
1
,
1
160
,
1
.
1
120
.
1
.
1
80
40
1
0
8 (ppm)
8 (ppm)
182
Problem 72
2000
3000
4000
1200
1600
800
V (cm
100-
Mass Spectrum
No significant UV absorption above 220 nm
c 5 h7 n o2 120
80
40
200
160
240
280
m/ e ....'.......1........'..... ..1.... —r ------ r................1....... ' ....... 1
1
I
•
I
»
I
1
'
1
•
'
r
_
i
1
i
1 ............
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f C H f
proton decoupled ■
i
i
i
I
i
i
200
i
i
i
i
i
160
I
solvent
1
i
i
i
i
i
120
i
. i
i____i____ i____ i
80
______ 8 (ppm)
__i____ i____ i
40
0
1H NMR Spectrum (200 MHz, CDCI3 solution)
r -
,---------j TMS
|
................................ .......................... — 1—
10
1—
1—
9
1—
1—
8
1—
1—
7
.
1____ 1____ 1...... 1
6
5
1
4
1\ 1
1
j
1
..... 1........ 1...... 1
3
2
1 .1
1
j
. 1
0 8 (ppm)
183
Problem 73
4000
2000
1200
1600 V (cm 1 )
3000
800
No strong UV absorption above 220 nm
120
80
40
160
200
240
280
m/e ■— | - 1
|
.
|
1
|
.
|
!
b
1
'
1
,
.
1
1
y— .
|
,
1
1
1
,
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
ch 2| ch 3| c h |
proton decoupled
solvent ill
I
--------- 1----- »----- 1----- 1----- 1----- *----- 1----- «------1----- »----- 1----- 1------1----- 1----- 1----- L
200
10
184
9
160
8
7
120
6
5
i
80
4
40
3
2
1
1
1
0
6
(ppm)
0 5 (ppm)
Problem 74
UV Spectrum
X max 283 nm (log10s 1.4) solvent: hexane
40
80
120
160
200
240
280
m/ e i — 1— r 13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
ch 2|
CH3f
ch|
proton decoupled solvent
* 1 I
J
1
I 200
1
I
1
L 160
_L 80
120
J
L.
0
40
8 (ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
4.0
J 10
1-----
3.0
exchanges with D20
2.0
ppm
1----- 1___ I___ 1____I___ i____> 8
6
■
l
■
l
■
l
1
I 1
0 8 (ppm)
185
Problem 75
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e 13C NMR Spectrum (100 MHz, CDCI3 solution)
D EP T
CH2^ CH3f CHf
solvent
proton decoupled
> «__ I__ 1__ I__ 1__ I__ 1__ I---- 1----1---- 1---- 1__ 1__ 1__ 1__ I__ i__ I__ 1__ I__ 1 200 160 120 80 40
I__ 1______ 0 § (ppm)
8 (ppm)
186
Problem 76
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e ,3C NMR Spectrum (100 MHz, CDCI3 solution)
DEPT
ch 2|
CH3f
ch|
solvent
proton decoupled _J
I
I
200
1
I
I
I
160
»
I
I
»________1
120
I________ t
i
80
l
l
!
1
________ I________ |________ I________ |________ ________ L_
40
0
5 (ppm)
5 (ppm)
187
Problem 77
2000
3000
4000
1200
1600
800
V (cm1)
100;
Mass Spectrum
80 No significant UV
60
absorption above 220 nm
40
20
M+* = 88 (< 1%) C5 H1 2 °
120
80
40
200
160
240
280
m/ e '
i
i
■r” ...r
1
i
1
1
i
1
i
i
1
1
i
i
1
i
1
1
1
i
1
1
1
r
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
CH2i|f CH3f C H f
proton decoupled
solvent^
____ 111 1 ____ 1 1 1
— 1------1------1------1------1------1------1----- 1----- «------1------1----- 1----- «------1----- •----- 1— 1
200
10
188
9
160
8
7
120
6
80
5
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
Problem 78
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e t—
1— I— •— r
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
CH2(i CH3f CHf
solvent proton decoupled
J
200
1
j
L
160
120
I
»
I
80
.
I
1
I
40
1
I
L
0
5 (ppm)
5 (ppm)
189
Problem 79
4000
3000
2000 1600 V (crrf1 )
1200
100;
800
Mass Spectrum
80 No significant UV
60 40
absorption above 220 nm
107/109
M+'
20
152/154
180/182
C5H9 ° 2 Br »
.
I
.
i
i___ ____ I___ ____ I___ ____ I___ i___ L .
.
40
80
120
160
200
240
280
m/ e i— ■ — I— 1— r 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
ch 2| ch 3|
CHf
solvent proton decoupled
J
.
I . 200
I
1
L 160
J
J
L.
120
.
I I 80
\__
I 40
-I
I 0
I—
8 (ppm)
6 (ppm)
190
Problem 80
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
CH2J CH3f C H f
proton decoupled solvent
J
I
I
i
I
I
I
I
I
I
I
.
I
,
120
160
200
I
.
I
I
80
I
L.
J
40
L-
0
5 (ppm)
i
1
1H NMR Spectrum (400 MHz, CDCI3 solution) expansion
5.0
J
10
expansion
4.5 ppm
3.5
3.0
2.5
ppm
1------- 1------- 1------- 1------- 1------- 1------- 1____ I____ 1____ >
8
1
1
■
i
■
i
»
1
0 8 (ppm)
191
Problem 81
4000
1200
2000
1600 V (cm'1 )
3000
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/ e 1
1 1 ' "I..... 1 13C NMR Spectrum
,
,
.
,
-|
.
|
.
,
.
|
.
•
1
>
1
1
1
1
1
•
1
.
1
'
1
1
(50.0 MHz, CDCI3 solution)
D EP T
c h 2|
c h :it
CH f
solvent proton decoupled
-
-1
-i— 1
L ............................................. • -...1 1 I -l 1 —1___ L..
200
10
192
9
160
8
7
1 1
1
120
6
5
1
80
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
Problem 82
4000
100
3000
:
60
. •
800
Mass Spectrum
69
41
80
1200
2000 1600 V (cm'1 )
^ CO CD
97
Q.
CD CO
No significant UV
115
r .Q 40
7 _
absorption above 220 nm
o M
+ ’ =
160
(<
1%)
20 L i
i
iiii
_________ i____1___ j ____i
It!
uU
_ i ____ i_ _ _ l
_
i
.
120
80
40
, ( 7 — 1------*-- 1------■------•------■----- «
*
c 7 h 12o 4 -- •-- L . 1 * 1 *
-- 1
<
*
-•------1
200
160
240
280
m/e 1
1
■
1
■
;
1
1
1
1
1
•
1
1
1
1
1
1
■1
r .•
■■■■■'
1 ........>
13C NMR Spectrum (125.0 MHz, CD3OD solution)
DEPT
CH2ijf CH3f C H f solvrent
proton decoupled
^| ................
I
1
1
I
•
1
200 1H
1
1
160
.
1
.
1
.
120
NMR Spectri
1
1
.
1
.
80
.
I
.
40
TMS 1 I
0
1
8
(ppm)
2 protons exchanged for deuterium in this solvent
(600 MHz, CD3OD s•olution)
expansion
1
r ‘ 2.2
1 1.8 ppm
solvent residual
solvent residual
i ----- 1--------- 1---------1--------- 1--------- 1---------1--------- 1---------1--------- 1---------1---------1.......... 1
10
9
8
7
6
5
4
1
■
3
r
____| ___
l
l
1
I
2
TMS
I
■
1
1
1
I
0 8 (ppm)
193
Problem 83 A Ir'
IR Spectrum (liquid film)
i
.
i
^
i
.
i
.
»
»
.
1640 / II « ■ I_______.___,___I_______I_______L
1200
2000
1600 V (cm'1 )
3000
4000
.
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e i
■
1
i
1 ......r "
i
1
i
i
•
1
i
1
i
1
i......1 ........ i
|
1
i
1-
13C NMR Spectrum (125.0 MHz, CDCI3 solution)
D EPT
c h 2|
c h 3|
ch
|
solvent proton decoupled
|
TMS
^
1 ------ 1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1------ 1
200 1H
160
120
1
1
1
1
80
1
1
1
1
40
1
0
5 (ppm)
NMR Spectrum
(600 MHz, CDCI3 solution)
TMS
1 1
10
194
*
1
9
1
1
8
1
1
7
1
1
J___1
6
5
.
i
4
1
.
3
.
2
1
.
1
1
1
1
0 5 (ppm)
Problem 84
1200
2000
1600 V (cm*1 )
3000
4000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e
200 1H
160
120
80
40
0
8
(ppm)
NMR Spectrum
(400 MHz, CDCI3 solution)
expansion
expansion
I 3.8
3.6
ppm
I
- 1------ 1— 1 ..T. 1.6 ppm 1.8
■ —
TMS
1 ----- 1--------- 1--------- 1--------- 1--------- 1--------- 1---------1---------1---------1---------1---------1---------1—
10
9
8
7
6
5
4
j L
1
1
3
1
I
2
L 1
I
1
1
1
1
0 5 (ppm)
195
Problem 85
No significant UV absorption above 220 nm
40
120
80
200
160
240
280
m /e .............1 ' 1 1 13C. NMR .^nAr.tri im
1
1 T " .... '
1
QEPT
(100.0 MHz, dioxan-D8 solution)
1
1 ”r
1
'
1
1
1
i
.
I
1....... 1
l
■
I
.
l
CH2| CH3f C H f
proton decoupled
I
.
I
.
I
.
I
.
I
.
160
200
i
.
.
.
I
40
80
120
.
.
I 0
. 8
(ppm)
1H NMR Spectrum (400 MHz, dioxan-D8 solution)
TMS
L I
10
196
i
1
•
I
8
i
1
i
—I
-i------- 1
—i
1
i
i_
1-------1—
1------- 1------- 1------- 1------- 1----
1
0 5 (ppm)
Problem 86
1200
2000
1600 V (cm"1 )
3000
4000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e 1...... '
1...
1 .. 1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
,
D EP T
.
I ch 2(i
.
1
I
1
1
'
1
'
I
-
1
'
1
'
I
.
I
«
1
CH3f CHf
solvent proton decoupled
I
■
I
u .
i
200
.............................................
160
i
120
•
80
40
I
i
0
8
(ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
TMS
I ---- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1— -L -
10
9
8
7
6
5
1
4
1
1
3
1
1
2
1
I
1
1
1
1
0 8 (ppm)
197
Problem 87
4000
3000
1200
2000 1600 V (cm"1 )
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e 1
------- r 1 i 1 .................................. ....... 13C NMR Spectrum
|
!
|
*
|
1 " 1..... 1
1
1
1........ ..
1,
1........._
(50.0 MHz, CDCI3 solution)
proton coupled
j. . . . .
j
ta.___________
_______ i
solvent proton decoupled
i
I .... 1..... 1
............. L_ .
1
I
200
I
1
ill
1
160
1
.
1
.4.. J..... 1
120
1
1 .. J ....... 1
80
1 ,1
40
0
5 (ppm)
1H N M R S p e c tru m
(200 MHz, CDCI3 solution)
r
expansion 1
l
with resolution enhancement
1 4.2
1............r.. 5.4 ppm
....I
5.8
10
198
...........i.. -
9
1
1
8
1
,
7
1
,
6
1.j V.
V
jL
I
expansion
with resolution enhancement
,
5
1
T
4.0 ppm TMS 1
.
I
4
...........
1
3
1
.
.....L........._L
2
1
1
.1_____L
0 8 (ppm)
Problem 88
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
n— t1— 1r
1----'---- 1----'----1---- T-------- '----1----1----1----r
r-
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
C H C H 3f CHf
solvent
proton decoupled
I
10
.
I
.
1
8
I____ 1____ I____ 1____ I____ 1____ I____ 1____ I____ 1____ I____ 1-------L
J
1
L_
0
5 (ppm)
199
Problem 89
No significant UV absorption above 220 nm
m/e
— r
1— 1— I— 1— I— r 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT
ch 24
CH3f
ch|
solvent proton decoupled
J
200
!
I I
I
I
160
I
L
-1
I
120
L.
J
I
1 I 80
I
I
I
40
I
L
0
S (ppm)
8 (ppm)
200
Problem 90
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
120
160
200
0
40
80
5 (ppm)
1H NMR Spectrum (300 MHz, CDCI3 solution)
expansions
Ji 3.8
3.7 ppm
2.1 2.0 ppm
r
1.3 1.2 ppm
TMS
J
10
1
I
i
I
8
i
I
■
I
6
A_________L -i__ i____ i_____i__i_____i__i_____i__ i i 4
3
2
1
0 5 (ppm)
201
Problem 91
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e 1
1
1
I
'
I
1
I
'
1
1
I
1
I
'
1
'
I
'
.
I
.
I
'
1
'
I
.
I
.
1
1
'
I
1
13C NMR Spectrum (125 MHz, CDCI3 solution)
D EPT
c h 2|
C H 3f c h |
solvent -fo
proton decoupled
I
,
I
200
.
1
.
1
160
.
I
.
I
120
.
1
.
1
80
40
0
5 (ppm)
8 (ppm)
202
Problem 92
100
55
Mass Spectrum
82
80 60
• -
No significant UV
CD CO 03
absorption above 220 nm
40
20
M+' = 83
C5H9N
— 1-1--.--1_1_I_1_I_1_I___I..... ............................... -I 1--1-- 1--.--1— 40
80
120
160
200
240
280
m/ e I ‘ I ' I 13C NMR Spectrum (100.0 MHz, CDCI3 solution)
'
1
I
I
'
I
1
I
'
I
1
I
1
I
1................
solvent ^
______ ______________ _______________________ I .................................................................... .......
200
1
I
CH2{ CH3f C H t
proton decoupled
.
I
|
D EPT
1
1
160
120
.
80
_ j— ,— 1— 1— 1— ,------------
40
0
5 (ppm)
6 (ppm)
203
Problem 93
100 ;
Mass Spectrum
43
80 No significant UV
60 40
71
absorption above 220 nm
M+’ = 86 (< 1%)
20 c 5Hi 0o _i
.
i
.
i
i
40
i— •— i— .— i— i— i— *— i—
120
80
160
200
240
280
m/ e -------- r
1
i
1... i
1
i
^
i
•—
i
•
i
'
1
t
r
1
nr
«
1
'
1
i
1
i
1
1
1
i
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
CH2i|f CH3f C H f
solvent
proton decoupled
111
----------- 1-------1-------1------j-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1_
200
10
204
9
160
8
7
120
6
5
i . l
i
80
4
40
3
2
0
1
8
(ppm)
0 5 (ppm)
Problem 94
1200
2000
1600 V (cm'1 )
3000
4000
800
UV spectrum 49.5 mg /1 0 mis path length : 1.00 < solvent: hexane
i . . . . . . . .. i .... 120
80
40
200
160
240
200
280
250
m/e 1
1
I
■
I
■
1
■
300
.
350
X (nm) 1
■
1
1
1
1
1
1
1
1
1
1
1
1
1
1 ....r-
1
1
1
1
1 —.....-
13C NMR Speictrum (100.0 MHz, CDC l3 solution)
D EPT
c H2(,
C H 3f
ch
|
proton decolupled solvent 1 1
I
1
1
1
1
200
1
1
1
1
160
1
1
1
1
1
120
1
1
80
40
1H NMR Spectrum
1
0
1
8 (ppm)
expansion scale expansions
(400 MHz, CDCI3 solution)
0
20
Hz
2.2
2.3
TMS
-J 10
1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1___ 1___ I___ 1___ I___ 1___ I___ 1 9
8
7
6
5
4
3
2
1
1
1
1
0 8 (ppm)
205
Problem 95
1200
2000
1600 V (cm'1 )
3000
4000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e ..I
1
I .... J
1
i
i
•
1
i
i
*
i
■
i
1
I
'
.
I
.
I
1
I,
.
I
,
I
,
,
.
13C NMR Spectrum (100.0 MHz, D2 O solution)
D EP T
ch 2|
CH3f CHf
proton decoupled
I
.
I
.
I
200
.
I
.
I
.
160
I
.
I
,
120
I
80
40
I
0
.
5 (ppm)
1H NMR Spectrum expansions
(400 MHz, D2 O solution) Note: there are 4 protons which exchange with the D2 O solvent
t-------1------ 1-------r 3.4 ppm
1.3
1.1
ppm
H2O and HOD in solvent
■1 J
10
206
9
8
7
1
I
L
6
5
■ 4
l
1 3
I
J
2
1
1
L
0 6 (ppm)
A-(nm)
m/ e I
1
I
.
I
.
I
.
I
1
.
'
1
'
1
1
1
'
1
1
1
'
1
'
R e s o lv e s into
13C NMR Spectrum
tw o sig n a ls a t h ig h e r field
(5 0 .0 M H z , C D C I 3 so lu tio n )
ch2\ C H 3f
D EP T
c h | i
R e s o lv e s into
i
r
tw o s ig n als a t h ig h e r field e x p a n s io n s _ I p ro to n d e c o u p le d .................................................................... ! I
1
1
1
1
200
10
9
1
1
1
..... 1
160
8
I
s o lv e n t
7
_-
.
........
»
1
1
. _
Ui 1
1
120
6
5
3 5 .0
1
l
______ L _
I
'
'
3 4 .0
ppm
_____ 1
1
1---------1--------- 1---------1--------- 1---------1--------- 1--------------------
80
4
i I
1
40
3
2
0
1
5
(ppm)
0 5 (ppm)
207
m/e T
X(nm)
'----1----'----1----'----I----1---- 1----1----1----1----1----'----1----'----1----'----1----1----1---- 1----1---- r
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
c h 2|
c h 3|
CHf
r- .
J
i
I
200
.
H
solvent
proton decoupled
I
i
I
160
i
I
r
i
. — .........
I
120
i
I
r .
i
I
80
____
___________ J
i
I
i
I
40
i
I
i
I
0
i________I 8
(ppm)
8 (ppm)
208
Problem 98
4000
2000
1200
1600 V (cm'1 )
3000
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
200
160
120
80
40
0
8 (ppm)
8 (ppm)
209
m /e |
1
|
—T —
1
|
|
,
X (nm) ,
|
.
|
.
|
i
, ....r "....r™... 1......i...
1......r......■.............
expansion
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
i .........1-------- 1— 140 130 ppm
i D EPT
CH2|
c h 3|
ch
|
expansion
I
140 proton decoupled
----- »----- 1----- •----- 1----- *----- 1----- 1----- 1----- •------1------1------1----- 1----- 1----- •
200
10
210
9
160
8
7
120
6
5
130 ppm ryv- solven't
1
i
I
i
I
80
4
i
I
40
3
2
i
l
0
1
i
5 (ppm)
0 5 (ppm)
Problem 100
UV Spectrum ? lmax 262 nm (log10s 2 .5) solvent: ethanol
40
120
80
160
200
240
280
m/ e |
r
|
,
|
,
|
,
|
,
l
'
I
1
I
I
'
I
.
I
.
I
,
l
.
I
.
13C NMR Spectrum (100.0 MHz, CDCIj solution)
expansion
i
-i------------------- 1— 130 125 ppm DEPT
ch 2|
CH3f CHf expansion
i
I 130
l
.
I
200
125 ppm r->- solvent
proton decoupled .
I
.
I
160
,
l
.
I
120
.
I
,
I
80
,
I
i .
I
40
0
5 (ppm )
5 (ppm)
211
Problem 101
100 r
Mass Spectrum
105
80 -■ CO - 0 0 60 "■ C - CO O -Q 40 7 o -
UV Spectrum ^max * 260 nm (log^s « 2.5)
00
II -i-
20 II 1 II __ 1_1_1___1_I_ 1 - 1 40 80
C 10H 14
lI
1 —1_1_._1_._1___1_1_I—* -«—* J * • *-- •-- *-- 1-- * 120 160 200 240 280
m/e 1
T..... 1 "1
1
1 "H..... '
1
1
'
1.. '
1
1
1
1 ... i
1
i
i
.■
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f C H f
1
1
solvent proton decoupled
t i
I ....
I
1
.
1
,
200
10
212
9
.
1
. .... 1
160
8
7
. 1 _ ,
1
,
120
6
5
1
,
1
1
,
80
4
1
,-
40
3
2
.
1
0
1
,
5 (ppm)
0 6 (ppm)
Problem 102
- 1 ...........
1 ------- ' ' ’ • 1 ................ ..
'
1.2
80
60
40
Mass Spectrum
119 % of base peak
' 1 ' i 1 t t't ' 1 ' 1 ' 1 »1 ' 1 ' 1
100
1.6
2.0
-
_
w 0 CD 0
\ \
""
\ \
UV spectrum
M+* 134
\
2.4
20
lr
I „ C 10H 14 ........... 1-- •---1-- •-- 1-- ■-- 1-- •-- 1-- --- 1-- .-- 1-- .-- 1-- .-- 1-- 1_1_1_1_._1_1_1_1_1_._1_1_ 200 240 280 120 160 40 80
1
m/ e
200
250
300
350
X (nm)
n— ■ — i— 1— r 13C NMR Spectrum (125.0 MHz, CDCI3 solution)
DEPT
CH2i|f CH3f C H f
solvent
proton decoupled
i__ _i
i
1
i 1 200
i
1
i 1 160
i
1
i 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1__ 1------120 80 40 0 5 (ppm)
5 (ppm)
213
Problem 103
4000
3000
2000 1600 V (cm"1 )
1200
800
UV Spectrum ?tmax 260 nm (log10s ^10 2.5) solvent: methanol
40
80
120
160
200
240
280
m/ e 1
1
1
■.. "1
1
1
1
1
J
1
.
1
.
1
1
.
1
.
1
.
1
,
. . . . 1. 1 .
I
1
I
1
1
.
1
.
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
ch 2i)i ch 3|
CHf
solvent 1
proton decoupled
t
1
.
1
.
1
200
.
1
160
.
1
.
1
.
120
1
.
1
80
40
0
s (ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
r
7.5
7.0
ppm
r .......... >
_ A1 ,
10
214
8
.............- __________ ______
____
L
TMS
-
_I
1
0 8 (ppm)
Problem 104
4000
2000
3000
1200
1600 V (cm-1)
100
800
Mass Spectrum
147 139/141
80 60 40
M+‘ 182/184
20 C11H15C| 40
120
80
160
200
240
280
m/ e ........i..... 1 ....... i”.... " r
"i
1—
i
1---- T ..
r
1—
13C NMR Spectrum
1
1
'
1
1....... 1....
'
1
1
1
................
Resolves into two signals at higher field strength
(50.0 MHz, CDCI3 solution)
D EP T
'...... I
ch 2| ch 3| c h |
Resolves into two signals at higher field strength
proton decoupled
^
solvent
^
III I
.
I
.
I
200
10
9
.
I
1
1
160
8
7
*
I
*
l
1
120
6
5
l
*
I
4____ l____1-------1-------1-------1-------1----------------
80
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
215
Problem 105
4000
3000
1200
2000 1600 V (cm"1 )
800
UV Spectrum ^ max 262 nm (log10e ^10 2.5) solvent: methanol
40
120
80
200
160
240
280
m/e 1 1 1 ' ... . 1 13C NMR Spectrum
■".
1 ■1
1
J
T
'
J
u
>
i
1
1
t
1
1
1
1
1
'
1—
1:15 ppm
130 CH3|
r
expansion
(50.0 MHz, CDCI3 solution)
DEPT CH2|
•
T
CH f
i expansion
J
L
130
......... I
....... I
proton decoupled I * I ....
I
216
9
I
1
1
160
200
10
.i
_
8
7
1:*
PPm
1- J_.. »
1
.
1
1
1
.
80
120
6
11
^ solvent
JO
_.
5
4
.
40
3
2
1 0
1
8
(ppm)
0 8 (ppm)
Problem 106
4000
1200
2000
1600 V (cm‘1 )
3000
100
800
Mass Spectrum
183/185
UV Spectrum
80: ? lmax 258 nm (log1Qs 4.2) 60 155/157
solvent: ethanol
M+’ = 198/200
40
20 C8H 7 O B r 40
120
80
160
200
240
280
m/ e i
1
1
i
i
1
1
r
1
i
1
J" “ 1
T
1
|
|
I'
|
'
|
P
i
.
|
1
i
I.....
13C NMR Spectrum (100.0 MHz, CDCIj solution)
1 D EP T
CH2(| CH3f C H f
proton decoupled
solvent
i i
.
i
i
.
i
.
i
200
.
i
1 .
160
1
. i
i
i
.
120
i
.
i
j
80
— i
40
0
S(PPm)
i
i
1H NMR Spectrum
J
10
1
I
9
i
I
8
i
i
7
i
l
6
i
5
i
i
4
i
i
3
i
i
2
i
i
1
i
0 5 (ppm)
217
m/ e ' 1 1 “1— 1 ” 1 13C NMR Spectrum
1
1
X (nm) 1
1 T
1
1 .1..... 1
.
|
'
|
1
I i 40
I
i
I 0
i 5 (ppm)
expansion
(100.0 MHz, CDCIj solution)
D EPT
r
'.. 1 .1
CH2-j[ CH3t C H f
140 135
ppm
expansion
I I
proton decoupled
“r....... r------ r~
I •
I
i
solvent-r|n
............... LI
l i 200
I
i
I i 160
I
i
I i 120
i
I
^
I i 80
PPm I
i
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
___ | l r~
—
8.0
1---7.8
I L_ ___ « 7.6
T"" 7.4
1 ppm
—*
TMS
_ _ ..
L
---- 1
10
1
1
9
—1—
J—
8
1------
7
1
1 ------- 1 ------- 1------- 1------- 1------- !
6
5
4
1
3
1
1
2
1
1
1
1 1
»
1
0
6 (ppm)
218
Problem 108
-1 rTT» T»"T ' I * 1' 1' 1' 1' 1
1200
2000
1600 V (cm'1 )
3000
4000
100
40
20
base peak
60
Mass Spectrum
119
UV Spectrum
^max 254nm (log^c 4.3)
1
^max 258 nm (log^s 4.2 )
% of
80
800
^ max 279 nm (log1Qs 3.3)
M+' = 154/156
^ ~ | C gHyO CI — i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i—.—i . i . i . i . i . 120 160 200 240 280 80 40
A,max 290 nm (log10e 3.0) solvent: hexane
m/e I
1
'
'
I
'
1
I
1
I
■
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
c h 2|
CH3f
1
r
1
1
■ 1
'
1
•
1
•
1
.—
1—
.--------------
expansion
!
|
132
|
ch
131
ppm
|
expansion
I
I
; i 1 proton decoupled
132
. i
200
i
131
ppm
.
i
solvent 1
I
L
..................................... .........
( 1. A
.
i
160
............................................... .........
120
.
80
40
0
5
(ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution) expansion
—
_
I j — i—
10
1—
1—
9
1—
1—
8
r~
__
8.0
i I 1—
7
_^
j 7.5
7.0 ppm TMS
1 1
1
.
6
.
5
1
1
4
1
,
3
1
,
2
1
.
1
1
.
1
0 6 (ppm)
219
Problem 109
100
Mass Spectrum
UV Spectrum
80 ^max 224 nm (log10e 3.4) 60 s o l v e n t : m e th a n o l
40
20 C8H8^3 »
n. ■
i
■.. 1
1
1
1
1
13C N M R Spectrum (50.0 MHz, CDCI3 solution)
D EPT
c h 2|
CH3f
ch
|
solvent proton decoupled
|
1
li
-------1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1--- 1-- l_
200
160
120
i
1
i
1
80
i
1
i
40
1
0
i
S(ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution) expansion with resolution enhancement
--------------- ».Ip---------------------- i ..... .....i.......
i
........... .
I i
160
I
I i
I
120
1
I
1
ill l
80
.
I
1
40
I
1H NMR Spectrum (200 MHz, CDCI3 solution)
s
i
TMS
1 j
10
8
L l_ 0
8 (ppm)
229
Problem 119
4000
3000
2000
1200
1600
800
V (cm-1) 135
100 cD 80 -■ ±C -. C D. Q 0/> 60 -■ < -. _C D Q 40 7 O
UV Spectrum A,max 262 nm (log10£ 2 .6 ) A,max 269 nm (log1Qs 2 .6 )
150
20 : :
Mass Spectrum
43
solvent: methanol
M + '= 192
i C i2 h 16o 2 ii , i. u ,i ___ _ I , i . i . i ,—i—.—i—.—i—,—i—,—i—.—i—,—i—,—i—,—i—,—i—.—i—.—i—,—i—,— 280 200 240 120 160 80 40
m/e I
1
1............... 1
1
1....... 1
1
1
1
|
.
|
1
|
|
1
1
|
1
1
1
1
1
l
*
0
8
13C NMR Spectrum MHz, CDCI3 solution)
(100.0
DEPT
CH2i|( CH3f CHf
1
expansion solvent
Iw ....r1.
proton decoupled
I
149 ppm
1
1 1 ----------- 1----- 1------1----- 1------1----- 1------1----- 1------1----- 1------1----- 1------1----- 1----- 1----- 1
1
I
200
10
230
9
160
8
7
120
6
5
l
1
l
80
4
»
l
40
3
2
1
(ppm)
0 8 (ppm)
Problem 120
4000
2000
3000
1200
1600 V (cm'1 )
800
UV Spectrum
A/max ~ 240 nm (l°g10s >4'°)
40
80
120
160
200
240
280
m/e 1
1
1
1
1
1
I
1
1
1 ...r
1
i " ' 1
1
i
1........ i ...... 1
i
..i ....... ■ "‘ i
1
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EPT
CH2| CH3f C H f
_ _ .....................................................
solvent
proton decoupled 1
i 1
1
.
1
.
■
200 1H
I
1
1
.
•
160
1________ i
.......................... .......... i 1
1
■
•
120
1
1
.
80
...-J____ ■----------
1
.
40
0
5(PPm)
expansior1
I
r----
8.0
I
1
1
■
NMR Spectrum
(400 MHz, CDCI3 solution)
1
1
1
7.Ej
PPM
TMS
I 1
i
1
1
1
1
1
1
1
1
--------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- ---------------- __________ __________ 1 . .................
10
1
9
8
7
6
5
4
I
1
3
...............
i .........- ..... 1................... 1
2
.............
1
1
I
1
1
............ .................... .............
0 5 (ppm)
231
Problem 121
4000
3000
2000
1200
1600
800
V (cm‘1) UV Spectrum
X max 227 nm (log1Qs 3.9) ^max 305 nm (log^s 4.0) solvent: methanol
120
80
40
200
160
240
280
m/ e ....
r..... 1
!" '
'r
1
1
" ■
i
•
1
1
1
1— r
'
7
1
r
1 - ... 1
1
1
1
• ......
1
1
■
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
...... _
1
.
CH2i|f CH3f CH f
D EPT
proton decoupled
solvent
j
1
1
i
----------- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1------1----- 1----- 1----- 1----- 1----- 1----- 1----- 1
160
2 0 0
1
1
80
1 2 0
1
1
40
0
5 (ppm)
1H NMR Spectrum (400 MHz, CDCIj solution)
expansion
—
, 1___ 1 8.0
7.0
---
ppm
TMS
- i
1
10
232
1
1
9
1--
1
8
1---
7
1------- 1------- 1------- 1------- 1------- 1------- 1---
6
5
4
1
3
1
1------ 1
2
1
1
I
1 ------ 1
0 6 (ppm)
X(nm)
m/ e ■
i
1— r
1
1
1
1
'
1
“— 1
1
1 ... 1
I..... '
1
1
1
1
1
1
1
1................
,3C NMR Spectrurr1 (100 MHz, CDCI3 soluti on)
D EPT
ch 2| ch 3| c h |
solvent
* proton decoupled
.
1
1
1
1
1
200
10
9
1
1
1
1
160
8
7
1
1
1
1
1
120
6
5
1 1
1
______________ ______ 1 — 1----- 1----- 1----- 1----- 1----- 1----- 1-------------
80
4
40
3
2
0
1
S(PPm)
0 5 (ppm)
233
Problem 123
UV Spectrum
40
80
120
160
200
240
^m ax 225
nm (log^g8 4 . 2 )
^m ax 250
nm (log^c 3 . 8 )
^m ax 293
nm (log^s 3 .2 )
^m ax 330
nm (log^s 2 .6 ) solvent: methanol
280
m/e
1---- 1---- 1---- 1-------- 1— |----'— T
i
1
r
i
r
1
,3C NMR Spectrum (100 MHz, CDCI3 solution) expansion
i
i
134
D EPT
i 132
CH2ijf CH3t C H f
i
ppm
expansion
Ju l proton decoupled
-i
i
134
JL 132
ppm
X 1----1----1---- 1---- 1__ i----1__ i__ i---- 1----1__ i__ i__ i__ i__ i__ i__ t i i 200 160 120 80 40
i
t
i 0
i______ S(ppm)
1H NMR Spectrum (400 MHz, CDCL solution)
TMS
■i 11 1i 1" 1i 11 11
10 ppm
J
10
234
I
8
I
1______ 1______ i______ L
1
0 5 (ppm)
Problem 124
2000
1200
1600 V (cm'1 )
3000
4000
800
UV Spectrum ^max 2 77 nm (log1Qe 4 .2) solvent: methanol
120
80
40
160
200
240
280
m/e I
1
1
^
I
I
'
'
I
T~" | ■ 1 - |
1
|
1
|
'" I
.
I
■
1
1
...'
1
■
I
.
I
.....
13C NMR Spectrunn (100.0 MHz, CDCIj solijtion)
D EPT
c h 2|
CH3f C H f solvent
proton decoupled
1 i
I
I
■
I
•
I
200
■
I
•
I
I
■
160
I
.
I
.
120
80
I
40
0
S(PPrni)
expansion
1H Nl\/IR Spectrum (400 MhHz, CDCI3 solution)
__ ,1^ i
7.65
i
___;i[__ r ----r
7.45
i---- i
6.85
i —r
6.65 ppm
TMS
I ---- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1
10
9
8
7
6
5
4
1
i
3
1
i
2
1
i
1
1
i
1
0 5 (ppm)
235
Problem 125
UV Spectrum log10c > 4
40
120
80
160
200
240
280
m/ e 1— > — i— «— i— 1— i— ■ — r 13C NMR Spectrum
expansion
(50.0 MHz, CDCI3 solution)
DEPT
c h 2|
CH3f
ch
85
|
solvent
200
10
236
expansion
85
proton decoupled
160
8
120
80
80 ppm
80 ppm
40
0
§ (ppm)
0 5 (ppm)
Problem 126
UV Spectrum ^m ax 235
nm (log^s 4 . 1 )
A.max 2 6 5 nm (log10£ 3 .0 solvent:
40
80
120
160
200
240
)
ethanol
280
m/e
8 (ppm)
237
X (nm)
m/ e I
1
I
1
1
I
1
1 ..
1
|
1
1
1....
1
1
1
1
1
»
1
1 ..T
1
•
1
1
1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2{ CH3f C H f
solvent
proton decoupled
.
------ 1------- 1-------1------- 1------- 1------- 1-------1------- 1-------«------- 1-------•------- 1-------»------- 1-------1—
200
10
238
9
160
8
7
120
6
5
1
»
80
4
1
40
3
2
1
0
1
S(ppm)
0 6 (ppm)
Problem 128
100 80 60 40
20 40
80
120
160
200
240
280
m/e I
'
I
'
I
'
I
1
1
'
1
'
1
'
1
'
1
1
1
1
1
1
13C NMR Spectrum (100.0 MHz, CD3SOCD3 solution)
D EP T
CH2| CH3f C H f
1— ~1- solvent
proton decoupled
I
1
I
|
|
i
1
200
_j----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1------1----- 1----- 1----- 1------1------1-------------
160
120
80
40
1H NMR Spectrum
0
5 (ppm) expansion
(400 MHz,CDCI3/C D 3SOCD3 solution)
expansion
mi
________________________
in___
r solvent
11
10
9
8
7
6
5
4
1
. I___________ i________ __________ j i i i i i i i i__ 3 2 1 0 8 (ppm)
239
Problem 129
100 ;
120
Mass Spectrum
UV Spectrum
80 ^m ax
312 nm (log^s 4.3)
60 solvent: methanol
40
20 C8 H9 N O 40
80
120
200
160
240
280
m/e 1
I
1
1
I
1
l
1
I
i
■ "7
I
1
1
1
■
T1
1
'
1
1
1
1
,3C NMR Spectrum (100 Mz, DMSO-d 6 solution)
ch2\
D EPT
CH3f
ch
|
solvent — r proton decoupled 1
1
1
| 1
I 1
1
200
1
1
1
1
1
................... .. 1
120
160
n
1
1
1
1
80
1
1
1
1
40
1 0
1
5 (ppm)
1H NMR Spectrum (200 MHz, DMSO-d 6 solution) Exchanges with D2 O
I
residual H20 in solvent
TMS
1 10
240
8
1
0 6 (ppm)
Problem 130
1200
2000
3000
4000
1600 V (cm 1 )
800
UV Spectrum A,max 246 nm (log10e 4.2) ■'max 280 nm (log^s 3.1) solvent: methanol
40
80
120
160
200
240
280
m/ e 1
1
.
1
.
“1
’
'
1
1
T""
r
1
1
'
”1... 1
.1
1
1
T
i
—1
1
1
j— 1 —1------
1
1
,3C NMR Spectrum (100 MHz, CDCI3 solution)
D EP T
c h 2|
CH3f
ch
f
solvent
* proton decoupled
I
jj
1
— 1----- 1----- i----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1 —t-
200
10
9
160
8
7
1
i
120
6
i
»
i
80
5
4
40
3
2
0
1
S(PPm)
0 5 (ppm)
241
Problem 131
UV Spectrum
X max 250 nm (log10s 3.1) ^max 287 nm (log^s 2.2 ) solvent: chloroform
40
80
120
160
200
240
280
m/e I 1 1 '" I 13C NMR Spectrum
1.... 1
'
■7.....T— 7— 1...... r
1
'.... 1
1
1
1 ..1.... 1.......r "
1
1
1
1
1
(50.0 MHz, CDCI3 solution)
DEPT
c h 2|
CH3f
ch
|
solvent proton decoupled
I
i
i
_____________1 . ----------- 1----- «----- 1----- «----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- «----- i
200 1H NMR
160
80
120
40
1
1
1 0
8
(ppm)
Spectrum
(200 MHz, CDCI3 solution) expansion
/vi, n 7.6 exchanges with D20 on warming
10
242
1---- 1--7.0
ppm TMS
1 1
0 8 (ppm)
Problem 132
4000
2000
3000
1200
1600 V (cm'1 )
800
100
UV Spectrum
80 X max 250 nm (log10e 4.1) 60 ^max 285 nm (log^s 3.6)
40
solvent:
20 40
80
120
160
200
240
ethanol
280
m/e
200
10
9
160
8
7
120
6
5
80
4
40
3
2
0
1
S(ppm)
0 8 (ppm)
243
Problem 133
UV Spectrum ^ max 257 nm (log10e *10t 4.3) solvent: methanol
40
80
120
160
200
240
280
m/e ...!
'■
I
'
I
i
•
■
i
1
*
■ •—
r
1
1
.... r
r
' 1
...."r
t
1...
1
1
'
13C NMR Spectrum Resolves into two signals at higher field
(50.0 MHz, CDCI3 solution)
D EPT
ch
2| CH3f C H f
proton decoupled
1
.
I
.
.
Hi
1 I
.
I
.
160
200 1H NMR
solvent
i I I
Resolves into two signals at higher field
1
.
I
I
.
.
80
120
I
.
I
.
1
.
I
40
0
. 8
(ppm)
Spectrum
(200 MHz, CDCIj solution) expansions
V- pI [ p \ —i---- 1--- 1— 4.6
4.2
3.8 ppm
M
TMS
J
10
244
8
1
i
L
0 8 (ppm)
Problem 134
IR Spectrum
1730
(CHCI3 solution)
1200
2000
1600 V (cm'1 )
3000
4000
800
0.597 mg /1 0 mis path length : 0.2 cm solvent: ethanol
120
80
40
200
160
240
280
200
250
,
1
1
I
1
I
-1
I
300
350
X (nm)
m/e 1
1
1
1
1
1
n----- 1
I .1 80
II
I.
1
'
1
1
1
13C NMR Spectrum (125.0 MHz, CDCIj solution)
D EP T
CH2|
c h 3|
ch
|
solvent
proton decoupled
j
|
...... .-......... 1 >
10
I
I .I 200
•
I
I
9
I
8
. 1I L I. 160
7
I
I
6
I
.
I .I 120 I
5
I
I
I.
4
I
3
I .I 40
I
I
2
I
1
I
.
II 0
I.______ 5 (ppm)
0 5 (ppm)
245
X (nm)
m/ e
I I 200
10
246
9
I
8
I
I I 160
7
I
6
I
I I 120
5
I
I
I « 80
4
3
« ■ I i 40
2
I
1
«
I 0
»______ 5 (ppm)
0 5 (ppm)
X (nm)
m/ e I ' I ' 13C NMR Spectrum
I
I
'
'
I
1
1
1
1
,
1
|
1
’
.
1
|
|
1
(50.0 MHz, CDCI3 solution) Resolves into two signals at higher field
D EP T
CH2jf CH3t
ch
|
Resolves into two signals at higher field proton decoupled
|
ifj- solvent
^ .
1
1
«
1
1
200
«------- 1-------•-------1------- 1------- 1-------»------- 1-------•------- 1------- 1-------1-------1------- 1------- 1-------1-------1------- 1------- 1----------------
160
120
80
—I----- 1----- 1----- 1----- 1----- 1----- 1----- 1------ 1
i
10
5
9
8
Ill
7
6
I
i 4
40
I
i 3
i
' 2
i
' 1
0
6
(ppm)
i
i
i
0 6 (ppm)
247
X (nm)
m/e
expansion
13C NMR Spectrum CDCI3 solution)
(100.0 MHz,
II
D EPT 135
CH2| CH3f C H f
130
I h
1i |l
ji
'
126
ppm
125
expansion
ppm
note expansion
expansion
solvent 135
proton decoupled
130
ppm
126
125
ppm JL
_i
i
i
L_
200
160
120
80
40
0
5 (ppm)
1H NMR Spectrum expansion
(400 MHz, CDCI3 solution)
TMS
j\Rj 10
8
1
0 8 (ppm)
248
Problem 138
1200
2000
1600 V (cm'1 )
3000
4000
. 1. , . \ . , i , 1111111111 % of base peak
100
800
Mass Spectrum
UV Spectrum
75
80
? lmax 262 nm (log10s 2.3)
60
solvent: methanol
40
M+’ = 166 (< 1%)
135
91
20
» ......I,I ... c 10 h 14o 2 --------,— i— ,— i— ,— i— ,— i— ,— i— ,— i— ,— i— ,— i— .— i— ,— i— .— i— ,— i— ,— i— ,— i— .— i— .
40
120
80
160
200
240
I
1 " r
280
m/e I
1
I
1
I
'
I
1
I
1
1
r
~i_ t ..... r
'i
1
r
1
.
I
.
1
1............
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
ch 2| c h 3|
CHf
solvent
proton decoupled
I
.
I
.
200
10
9
in I
.
I
.
I
160
8
7
.
I
.
I
.
120
6
5
I
-
I
.
I
80
4
40
3
2
I
0
1
5(ppm)
0 8 (ppm)
249
Problem 139
1200
2000
1600 V (cm'1 )
3000
4000
800
UV Spectrum ? lmax 269 nm (log1Qs 2.7) X max 263 nm (log10e 2.7) solvent: methanol
m/ e --------------T”.. '
I
" '
l
'
l
1
'
1........1.......1...... T ......'........r
I
<
1
11
1
1
1
1............
I
.
I
.
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
proton coupled
..
.... .
.....
.......
i,
soh✓ent
proton decoupled I
,
I
| ,
I
| .
I
200 1H
.
I
__________ _____ ____ j I . I . I
.
160
120
.
I
_____ . I
80
.
40
0
5 (ppm)
NMR Spectrum
(200 MHz, CDCI3 solution)
expansion
1
1
2.5
2.0
1
1
1.5 ppm
TMS
__ . ... J I
1 ,, 1
10
250
1
1
9
1
1
8
1
1
7
•
1
6
1-
5
1 -
1---------1----------1--------- 1----------1----------1----------1----------1----------1----------1-----
4
3
2
1
0 8 (ppm)
Problem 140
! \
UV spectrum 0.491 mg /1 0 mis path length : 0.2 cm solvent: ethanol
40
120
80
160
200
240
200
280
250
1
'
1
1
'
1
1
300
350
X(nm)
m/e 1
1
1
■
1
1
1
1
1
r
1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
CH2(f CH3f
ch|
solvent proton decoupled
I
1 --- 1--- 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1_ .J_ _j___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1_______
200
160
120
40
80
0
5 (ppm)
1H NMR Spectrum (400 MHz, CDCIj solution)
expansion
expansion
-i-------- 1-------- 1— 4.6 4.4 ppm
_l
1
1----------
1.6
1.4
ppm
TMS
-i 10
1----- 1----- 1----- 1----- 1----- 1___ I___ i___ I___ i___ I___ i___ I___ i___ I___ i 8
I
■
1
0 5 (ppm)
251
X (nm)
m/e 1 -~r
I
-1
1
1
1
1
1 —.
.
|
|
-r
1
|
|
,
1
1
|
,
,
1
1
1
1
1
1
1
1
1
13C NMR Spectrum ( 100.0 MHz, CDCI3 solution)
D EPT
c h 2|
CH3f C H f
solvent proton decoupled
1 ----- 1------1----- •------1----- 1------1----- 1----- 1----- 1------1----- 1------1----- 1----- 1----- 1----- 1
200
160
120
1
80
40
0
5 (ppm)
1H NMR Spectrum (400 MHz, C D C I3 solution) expansion
expansion
expansion
—1------ 1
7.8
7.6
4.4
ppm
4.2 ppm
1.4
r
1.2 ppmi TMS
__ L
10
252
9
8
7
6
5
4
3
2
1
0 5 (ppm)
Problem 142
100 80 60 40
20 40
80
120
160
200
240
280
m/e --------- 1........1
I
1
I
1
I
1
I
—1
1
'
T - '
1
(50.0 MHz, CDCI3 solution) 135
D EPT
T"
■
I
1
I
1
I
1
1
1
I
1
I
< I
1
'
expansion
13C NMR Spectrum
1
i
130
PPm
CH2(f CH3f C H f expansion
!L
11
135
... . 1
1
130
ppm
solvent proton decoupled
..- ....... L_ j
|
I ... .
I
200
10
9
1
1
.1
I
160
8
11
A
7
1
1.
1
I
<
120
6
5
I
<
80
4
40
3
2
0
1
8 (ppm)
0 8 (ppm)
253
Problem 143
UV Spectrum
^max 220 nm (log^c 3.7) ^max 274 nm (log10s 3.3 ) solvent: ethanol
m/e
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
c h 2'|(
CH3f
ch
|
solvent
I
proton decoupled
j
1
i . 200
i
1
i 1 160
i
1
i , 120
i
■ 1 80
.
i
-I
I
40
L.
J
L
1
0
8
(ppm)
8 (ppm)
254
Problem 144
1600 V (cm'1 )
100 80 60 40
20
1200
2000
3000
4000
163
800
Mass Spectrum
UV Spectrum
- ^ ■ CD . Cl
? lmax 223 nm (log1Qs 3.9)
■ C/5 _ . -Q
^max 275 nm (log^s 3.1 )
0 0 0
“
‘o
:
^
[
I
:
it
i i
.
i
i
i
.
i
i .
» i
> i
.
I.
i i
+ '=
1 9 4
i
i
X max 281 nm (log10s 3.0 )
c 10h 10o 4
I.
i
120
80
40
M
135
1 0 4
.
i
.
.
.
200
160
a
.
i
i
solvent: ethanol
i
280
240
m/e ....1—
1
1
1 ......1....” T"
1
1
1
1
111
1
!...... r....... 1...... T
1 ... 1
1 .... r ".. 1
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
CH2iji CH3f C H t
solvent proton decoupled
*
i
1 I
.
!
.
I
200
10
9
.
I
.
I
160
8
7
.
I
.
I
.
120
6
5
I
.
I
1
1
80
4
.1.. -
40
3
2
1
-L----- 1----- i------------
0
1
8 (ppm)
0 8 (ppm)
255
Problem 145
4000
3000
2000 1600 V (crrf1 )
1200
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
I
I
I
m/e
1
1
I
1
T
I
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
c h 2|
CH3f C H f
solvent
proton decoupled
I I__ J
200
160
1
L
120
80
40
0
8
(ppm)
8 (ppm)
256
Problem 146
100 : -
60
-
CD CO CO o
7
o
800
Mass Spectrum
91
80
40
1200
1600 2000 V (cm'1)
3000
4000
UV Spectrum
CO CD
^m ax
261 nm (log1Qe 3.5)
M + '= 92
solvent: ethanol
20
c 7h 8 120
80
40
200
160
240
280
m/e I
1
I
1
I
1
I
1
I
T"
1
1
1
1
1
1
1
1
1
1
1
1
1
'
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2(( CH3f C H f
solvent proton decoupled
I
.
I
.
I
200
10
9
.
I
.
I
160
8
7
.................... I « I_ .
120
6
5
t i I
. ..J___-•___ I------.------1----- 1------1----- *------------
80
4
40
3
2
0
1
5
(ppm)
0 6 (ppm)
257
Problem 147
4000
3000
1200
2000 1600 V (cm’1 )
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
I
1
I
1
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2ijf CH3t
ch
|
proton decoupled solvent i
I
.
I
.
I
.
I
200
J
160
120
i
L
80
40
0
8 (ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
J
10
258
i
I
9
i
I
8
i
I
7
i
I
6
i
5
I
i
4
I
i
3
L
2
1
0 8 (ppm)
1200
2000
1600 V (cm'1 )
3000
4000
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
i
i
m/ e i
r
r
•
i
1
i
•
i
1
r .. 1
i
>
i
i
«
1
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2i|( CH3f C H f
solvent
1 proton decoupled 1
■
1
^ -
1
200
10
9
^ .
1
•
1
160
8
7
•
1
«
1
«
120
6
5
1
«
1
.
1
80
4
«
1
40
3
2
-i
-L.... .
0
1
8
(ppm)
0 8 (ppm)
259
Problem 149
4000
3000
40
2000 1600 V (cm'1 )
120
80
160
200
1200
240
800
280
m/e i
1
i
1
1
r
1
1
f"
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2\f CH3f C h {
proton decoupled
solvent
I
8 (ppm)
260
Problem 150
100
. M + '= 128
80 -' 60
. -■ _
40
20
. 7 :
Mass Spectrum
03 CD CL
No significant UV
CD C/3
03
-Q 0 ^8 0
absorption above 220 nm 55
83 ,
100
1
C 7H 12O 2 , 113 i ---- 1— 1— I— 1— I— 1— I— 1— I— •— I— 1— I— 1__I_____1__1__1_____1 . 1 . 1 . 1 . 1 . 1 .
40
80
120
160
200
240
280
m/ e l
1
r
1
i
1
r
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
c h 2|
CH3f C H f
solvent proton decoupled
5 (ppm)
261
Problem 151
4000
2000
3000
1200
1600 V (cm'1 )
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
i — 1— i— «
i
1
i
1
i
1
i
1
"1
r
'
1
1
13C NMR Spectrum
1
*“
expansion
(100.0 MHz, CDCI3 solution)
DEPT
c h 2J
CH3f
ch
1-------- 1------- r 28 27 ppm
|
expansion solvent
proton decoupled
1-------- 1------- 128 27 ppm
I - I
J
1
I . 200
I
1
I 1 160
I
1
I 1 120
I
1
I 80
....
1
I
.
I 40
1
I
1
I 0
1_____
5 (ppm)
5 (ppm)
262
Problem 152
1200
2000
1600 V (cm"1 )
3000
4000
800
No significant UV absorption above 220 nm
40
120
80
160
200
240
280
m/e I
1
1
1
1
■1— r .... r "
1
1
•“
1
■
1
1
'
1
1
1
1
1
'
1
1
,
1
!
0
5 (ppm)
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
CH 2| CH3t
ch
|
solvent proton decoupled
_____ L... 1
1
1......j__... 1,
200
10
9
1
,
1
,
160
8
7
—
1----- 1— 1
120
6
j— 1
1
80
5
4
1
.
40
3
2
1
0 5 (ppm)
263
Problem 153
100
:
80
-
■ -
03 CD
60
~ • -
CD CO 03 o
7
o
40
1200
2000
1600 V (cm'1 )
3000
4000
800
Mass Spectrum
30
No significant UV absorption above 220 nm 56
“
20
M + ‘ = 1 0 2 (< 1i
L
__ i___ i___ i___ _
>iJi i
i i, •
1% )
C 5 H 14N 2
,1
................................................
i ___ 1___ 1___ ____ 1___ .___ 1___ i----- 1----- 1----- 1----- .----- L.
120
80
40
200
160
240
,
.
280
m/ e I
1
1
I
i
■
i
■
I
1
1
'
1
1
1
1
1
1
•
1
1
■
1
13C NMR Spectrum r 'in n
DEPT
c h 2{
CH3f
ch
|
solvent if
proton decoupled
1 I
1
I
1
I
200
10
264
9
1
I
1
160
8
7
1
I...._ J ___1
1
1
120
6
5
1
1
..... 1....... 1........L ..... 1 ,, 1........1 ...... 1,.... JL.................
80
4
40
3
2
0
1
8 (ppm)
0 8 (ppm)
Problem 154
100 :
Mass Spectrum
31
80 ’ CO
: o-
No significant UV
0j) 60 ■< 40
~-Q 7o -^
absorption above 220 nm
M+' 100 < 1%
20
c 2h 3 f 3 o
^ IJ . iJ ... i _____i_i_i_i_i_*—
i— *— • * » • i — i— . i_j_1_1_1_1_1_1_1_._1_._1_._ 280 120 160 200 240 80
40
m/e ...1........1
1
1
1
" 1
I
1
1
J
"I
1
I
1
"1........r
1
1
1
1...... r
1
-l
T
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
— —
ch2| CH3f chf 1 1
V
I |
v
61.5
expansion
1 60.5 ppm
expansion l
solvent
proton decoupled
I
!i
— i------1------1— 61.5
J
........................................................................................I________I________ I________ I________ I________ L_
200
10
160
8
120
60.5 ppm
80
40
L-
0
1
8
(ppm)
0 8 (ppm)
265
Problem 155
100
:
80
-
60 40
119
■ “. -Q -
CO CD
-
CD CO CO
Mass Spectrum
UV Spectrum
91
log10s > 5
M+‘ 226
7 0
20
| , I C 15H 1 4 ° 2 I „ l —--L-.-- .--1--.--1--.--1--.--1--.-- --.-- --.-- --.--1--.--1--.--1--.-- --.-- --.-- --.-L
1
40
80
120
1
160
1
1
200
1
240
1
280
m/e
i
1
r
13C NMR Spectrum expansion
(150.0 MHz, CDCI3 solution)
DEPT
CH2| CHat C H f
130
129
128 ppm
expansion
130 proton decoupled
129
J
200
128 ppm
_____________solvent -rj~i
160
120
I
I
80
I
I
I
I
40
L.
J
L-
0
5 (ppm)
8 (ppm)
266
Problem 156
1200
2000
1600 V (cm'1 )
3000
4000
100 :
105
800
Mass Spectrum
80 -• 00 - CD
log10e > 5
. Q.
60 -
UV Spectrum
0
(/)
- CO . _Q
M+’
40 • ° -^
226
77
20
121
1
n
U
I I C 15H 1 4 ° 2 .i... i 1]__iLJ I s.l ------- .— i—.— i—.— i—.—j _i_i_ --- 1------- 1---.--- 1---1--- 1 _1_1_1_1__1_1_1__ » - » ■ ! . 40
80
120
160
200
240
280
m/e
200
10
9
160
8
7
120
6
5
80
4
40
3
2
0
1
5 (ppm)
0 5 (ppm)
267
40
80
120
160
200
240
280
200
250
I
1
1.
1
1
1
1
1
300
350
A,(nm)
m/ e 1 —r
1
1
|
1
|
1
|
1
1
1
I
'
1
*
I
1
1
1
expansion
13C NMR Spectrum (75.0 MHz, CDCI3 solution)
DEPT
135
CH2^ CH3f C H f
125
ppm
expansion solvent
-r-1 il 1 T 1 --- r- T---- T.. —
proton decoupled
135
...i_____________ L______ i ----- 1----- 1----- 1------1----- 1----- 1----- 1----- 1-----j_----- 1------1------1----- 1
200
10
268
9
160
8
7
l
1
120
6
5
l
i
125
t
1
1
80
4
ppm
1
40
3
2
0
1
5 (ppm)
0 5 (ppm)
40
80
120
160
200
240
280
200
250
T
'
I
'
I
1
I
1
300
350
A (nm)
m/ e I
'
1
1
r.....
1
i
1
r
1..
i
.nr"
I
•
i .. 1
1
r
expansion
13C NMR Spectrum (100.0 MHz, CDCIj solution)
D EP T
... f
CH2-If CH3f C H f
140
13-0
— T
ppnn
exparision solvent 11
— t
T........ 1 -.. T
proton decoupled
I
.
I
i
*
l
200
10
9
i
I
i
I
•
160
8
7
13 0
140
i
I
i
l
pprn
>
120
6
5
I
•
l
•
80
4
I
40
3
2
•
l
0
1
>
5 (ppm)
0 5 (ppm)
269
Problem 159
2000
1200
1600 V (cm'1 )
3000
4000
800
UV Spectrum l°9io e > 4
40
80
120
160
200
240
280
m/e
1----1 1
1
I
1
I
r
n
1
i
1
i
1
i
-t
n
|
,
expansion
13C NMR Spectrum
r
expansion
(100.0 MHz, CDCIj solution)
D EPT
CH2j CHst
ch
|
128.7 128.4 ppm
130
expansion
120
ppm
expansion • solvent
proton decoupled 130
128.7 128.4 ppm
1----- 1------1----- 1----- 1----- 1------1----- 1----- 1----- 1___ I___ i___ I___ i
-i
200
10
270
9
160
8
7
120
6
5
I
i
I
80
4
■
I
40
3
2
i
120
I
0
1
ppm
i
S(ppm)
0 5 (ppm)
Problem 160
UV Spectrum log10s > 4
80
120
160
200
240
280
m/e
5 (ppm)
271
Problem 161
4000
3000
2000 1600 V (cm'1 )
40
80
120
160
1200
200
800
240
280
m/e I
1
...1
1
1
1
CH3f
ch
1
1
1
1 ........r "....1
'
'
1
1
1
' ......T ” ....'
1
'
1
1
1
1
1
1
------
13C N M R S p e c tru m (50.0 MHz, CDCI3 solution)
DEPT
c h 2|
|
proton decoupled
solvent
| 1
it
----1-------•-------1-------1-------1-------•-------1-------1------1-------1-------1-------1-------1------J------ 1
200
160
120
L
1
80
1
40
0
1
5 (ppm)
1H NMR Spectrum (200 MHz, CDCIj solution)
expansion
expansion scale
8.0
8.5
10
272
9
8
7
6
5
4
ppm
3
2
1
0 S (ppm)
Problem 162
1200
2000
1600 V (cm'1 )
3000
4000
100 ;
800
Mass Spectrum
119
UV Spectrum
80 ^m ax
246 nm (log^s 3.7)
60 ^max 273 nm (log^c 2.6 ) 40
Mh
20
solvent: ethanol
191
c 12h 17n o 40
80
120
200
160
280
240
m/e
200
10
9
160
8
7
120
6
5
80
4
40
3
2
0
1
5 (ppm)
0 6 (ppm)
273
Problem 163
1200
2000
1600 V (cm'1 )
3000
4000
800
UV Spectrum
I
max
277 nm (log^s 3.4) solvent: methanol
120
80
40
200
160
240
280
m/e 1...
1
I
1
I
'
1
'
1
1
1 .... 1...
1
1
1
•
1
1
1
1
1
r
1
1
1
1
1
1
1
1
»..
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
c h 2j c h 3| c h |
solvent proton decoupled
|
in
--------------1-------1-------1-------1-------1-------1-------1-------•------- 1-------»------- 1-------»------ 1
200
10
9
160
8
7
j—
120
6
5
1
80
4
40
3
2
0
1
5 (ppm)
0
8 (PPm)
274
X (nm)
m/e
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
CH2| CH3f
ch|
solvent
expansion expa
jl
T------------ 1-------------T 175 170 165 ppm
1
proton decoupled I
»
I
I
I
200
10
9
.
I
J
L.
160
8
7
120
6
5
i
I
i
I
80
4
i
L
40
3
2
0
1
8 (ppm)
0 8 (ppm)
275
A,(nm)
m/e --------
I
'
I
'
I
'
I
1
T
1
i
1
|
■
|
.
|
,
|
.
|
i
|
.
i
1
i
1
i
1
i
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f C H f
r
solvent
proton decoupled
----- 1------1----- 1------1----- 1----- 1----- •----- 1----- 1------1----- «----- 1----- 1----- L
200
10
276
9
160
8
7
i
120
6
5
1
i
1
80
4
40
3
2
0
1
5 (ppm)
0 6 (ppm)
m/e I
.
I
.
I
.
I
■
X(nm) 1
“I
1
I
i
' 1
1
1
1
I
i
.
i
T
1
1 T
.
i
.
..... 1
13C NMR Spectrum (150.0 WIHz, CDCIj solution)
D EP T
CH2| CH3f C H f
solvent proton decoupled
I !
.
i
i
.
i
.
200
.
i
1
I .
i
.
160
i
.
i
.
120
i
1 .
80
40
i
0
.
S(ppm)
1H NMR Spectrum (600 MHz, CDCI3 solution)
Expansion Scale
exchanges with D20
7.00
6.98
6.72
6.70
6.68 ppm
vertical expansion x20
TMS
_L J ---------------- 1---------------- 1---------------- 1---------------- 1---------------- 1___________I__________ I___________I__________ I__________ I__________ I__________ I__________ I__________ I__________ L_
10
9
8
7
6
5
4
3
2
1
0
5 (ppm)
277
Problem 167
100
1200
2000
1600 V (cm-1)
3000
4000
■ -
60
* -
-
Mass Spectrum
261/263/265
:
80
800
182/184 CO CD 0 C/5 CO o
M+*
197/199
40 7
276/278/280
o _ -o
20
C9H10 Br2 _l
40
i
I i
80
I
I
L_
120
_l
160
i
I
200
.
I
.
I
I
240
■
t
I
u
L_
280
m/e I
1
I
'
I
'
1
'
I
'1
1
T_
1
1
T .. '
1
'
1
1
1
1
1
■
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
L D EP T
c h 2|
c h 3|
ch
1
|
solvent proton decoupled
i
.............111 .
______ 5 (ppm)
________ I____ 1____ 1____ 1____ 1____ 1____1____ 1-------1-------1-------1-------1---------------1-------1-------.-------1-------1-------.-------.-------.____ .____ 1____ 1
200
10
278
9
160
8
7
120
6
5
80
4
40
3
2
0
1
0 6 (ppm)
Problem 168
i ■» . ■■■• ■- I - i ■« ■t • • ^*
4000
3000
i— i
2000
i
i
t
— i— I— i— i— •— i—
1600
1200
800
V (cm'1) 100 80 60 40
20
m/ e i
i
'
1
1
r
r
I
(50.0 MHz, CDCI3 solution)
DEPT
'
one signal even at high field
13C NMR Spectrum
c h 2| CH3f c h |
Resolves into two signals at higher field
solvent proton decoupled
1____ 200
10
9
J_ _ L
J
160
8
7
6
5
4
I
I
3
2
I
I
0
40
80
120
1
L.
5 (ppm)
0 6 (ppm)
279
m/e I
I
I
I
I
1
I
1
A,(nm) 1
I
1
1
1
1
1
1
1
'
1
1
1
1
,
1
.
1
1
'
13C NMR Spectrum (150.0 MHz, CDCI3 solution)
DEPT
ch 2| c H3f C H f
solvent proton decoupled
-
-L
1
■
1
1
-L
200
■
1
1
1
160
■
.....................................................1
120
80
40
1H NMR Spectrum
.
0
5 (ppm)
expansions
(600 MHz, CDCI3 solution)
TMS
Expansion Scale 0
10
280
9
8
7
6
5
4
3
5
10 15 20 Hz
2
1
0 5 (ppm)
Problem 170
4000
3000
2000
1600
1200
800
V (cm'1)
40
80
120
160
200
240
280
m/e
1
.
1
1
1
1
'
1
....1
1
1
'
1
1
1
1
1
1
1
1
'
1
1
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
CH2(f CH3f C H f
proton decoupled
1
1
......... ....... .............. . 1 . 1 .
200
......I ..... J 1 . 1
160
I .
solvent
L----1 .
.
.. 1
.
120
. 1
1
______ . 1
____ 1 .
80
1
40
.
, ...
1
0
8 (ppm)
1H NMR Spectrum (2 0 0 M H z , C D C I 3 so lutio n)
R e s o lv e s into tw o sig nals of e q u a l intensity a t h ig h e r field
7 .0
7 .5
p pm
TM S
_______«UUL _1_________ I_________ I_________ I_________ L .
10
9
8
7
6
5
4
3
2
1
0 8 (ppm)
281
Problem 171
3000
4000
1600
2000
1200
800
V (cm‘1)
40
80
120
160
200
240
280
m/e
200
10
282
160
8
120
80
40
0
6 (ppm)
0 6 (ppm)
Problem 172
4000
3000
2000 1600 V (cm"1 )
1200
800
m/ e i
1
i...~“r ..T "
i....r
*
1
T " T ....T...... r
1 ... i
1
i..
1
i
1
1
,
i
1.............
1
i
13C NMR Spectrum (50.0 MHz, CDCI3 solution)
DEPT CH2 |
CH3|
CH f solvent
proton decoupled
1
1
.
1, I
.
200
.
I
1
II,
.....................................1
160
1 , 1
,
120
1
,
,............
80
40
0
...
5 (ppm)
1H NMR Spectrum (200 MHz, CDCI3 solution)
1 10
9
8
7
6
5
4
3
2
1
0 8 (ppm)
283
m/e
200
X (n m )
160
120
80
40
0
8 (ppm)
1H NMR Spectrum (100 MHz, CDCIj solution) 6H
TMS
-IH
residual CHCI3 insolvent
1H
JL 10
284
9
8
7
6
5
4
3
2
1
0 8 (ppm)
Problem 174
100
68
Mass Spectrum
UV Spectrum
80
A,max 225 nm (log10s 3.9)
60
M+ '=
40
s o lv e n t:
96
methanol
20 c 6 h 8o _J
I
I
I
40
I
I
I
1
I
I
120
80
I
I
I
I
I
I
I
.
I
■
200
160
I
*
I
.
240
I
.
I
280
m/e
I 1 I 1 13C NMR Spectrum
I
1
I
1
1
1
|
1
|
1
|
1
|
1
|
1
,
,
,
.
( 100.0 M H z, C D C I3 solution)
DEPT CH2| CH3f CHf
1
proton decoupled
______ I
»
solvent
i I
200
1
1
1
* ................. 1 . ____ 1 *___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1-----1-----1-----1-----1-----1-----1---- 1---------0 8 (ppm) 40 80 120 160
8 (ppm)
285
Problem 175
2000
3000
4000
1200
1600
800
V (cm‘1 ) UV Spectrum ^max 263 nm (log^s 3.9) X max 305 nm (log10e 2.1 ) solvent: hexane
40
80
120
160
200
240
280
m/e ■............... I”
1
1
r .. r ™...i
..................................................
13C NMR Spectrum
1_ .....r — «"
expansion 40
c h 2|
CH3f
ch
1
"1.......1
1 ....
1
1
1
i
(100.0 MHz, CDCI3 solution)
DEPT
1
i i t—i—r 30
f
T—1--1--1--1— 20 ppm
| expansion
i —i—i
-r 30
40
proton decoupled
20 ppm
solve
nt 1. •
»
1
200
»
1
»
1
160
>----- 1---- ‘
i
120
»
>
'
l
80
i____ i
l
i
l
40
i
I
i
I
0
i
6 (ppm)
5 (ppm)
286
Problem 176
UV Spectrum ^max 234 nm (log^c 4.1) solvent: ethanol
40
80
120
160
200
240
280
m/ e
6 (ppm)
287
Mnm)
m/e '
I
1 “T
1
n— 1
1
1
1
r
1
’
1
'
1
1
1
'
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
l3C NMR Spectrum (100 MHz, CDCI3 solution)
D EPT
c h 2|
c h 3|
ch
|
solvent proton decoupled
*
--- 1----- 1----- 1------1----- 1----- 1----- 1------1----- 1----- 1----- 1------1----- L-----L
200
10
288
9
160
8
7
1
120
6
1 1
80
5
4
40
3
2
0
1
8 (ppm)
0 8 (ppm)
Problem 178
4000
2000
3000
1200
1600
800
V (cm'1) 100
UV Spectrum
80
^"max 260 nm (log^s 2.9 )
60
^max 266 nm (log^s 3.1)
40
X max 272 nm (log10s 3.1 )
20
solvent: methanol
40
80
120
160
200
240
280
m/e I
.
1
I
,
.
I
.
I
.
I
■
1
I
I
1
I
.
I
'
I
1
I
1
I
1
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
c h 2| c h 3! C H f
solvent proton decoupled
I
I I
.
I
■
I
200
10
9
•
I
.
«
I
160
8
7
.
I
.
I
.
120
6
5
I
.....i..
____ I____ .____ I------- .--------------
40
0
80
4
3
2
1
8 (ppm)
0 8 (ppm)
289
Problem 179
1600 2000 V (cm1)
3000
4000
1200
800
100
UV Spectrum
80 ^m ax
=
243 nm (Iocj^qS 4.1)
60 ^max = 291 nm (log1Qs 3.4) 40 solvent: ethanol
20 120
80
40
200
160
280
240
m/ e t—
■ — r
expansion
13C NMR Spectrum (100.0 MHz, CDCIg solution) 128
DEPT
CH2j CH3f CHf
126
124
ppm
124
ppm
expansion
128
126
proton decoupled solvent
J
1
I
I
1
200
10
290
9
1
I
1
I
160
8
7
.
I
1
I
1
120
6
5
I
I
1
I
I
1
80
4
I
1
40
3
2
.
L
0
1
5 (ppm)
0 8 (ppm)
"i « rm f r ' - ... 1-- ■ -- |-- --- i-- --- |-- ■ -- r » |--
| 1i ' » ' i ' i ' | '
IR S p e c tr u m
r
Problem 180
1751
(CHCI3 solution)
i • » * ■ • ■• » • I • ■ - ■ • ■ ■- I »_»___ 1.i * 1 ____1____L 800 1200 1600 2000 3000 V (crrf1)
4000
UV Spectrum
X max 268 nm (log10s 3.1) A-max 275 nm (log1Qs 3.1) s o lv e n t: ethanol
40
80
120
160
200
240
280
m/e i
'
i
1...“ i
1
i
-T—
r — r'
1
I
1
I
I
1
I
.
______ I .
.
I
1
I
1
I
.
I
.
I
.
I
..X.
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2| CH3f C H f
proton decoupled i I
I
.
,
solvent ..................................... I_____ , I , I . I .
200
10
9
160
8
7
1 I
.
I
.
120
6
5
I
80
4
40
3
2
0
1
5 (ppm)
0 8 (ppm)
291
Problem 181
4000
3000
2000
1200
1600
800
V (cm'1) UV Spectrum ^max 249 nm (log10s 4.1) ^ max 292 nm (log^s 3.3) solvent: ethanol
40
80
120
160
200
240
280
m/e
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
c h 2| CH3t c h |
ppm
130
125
expansion
proton decoupled solvent ppm
130
125
_L
200
160
120
J 80
I 40
I 0
8 (ppm)
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
] 10
292
8
lui.
TMS
jj 1
0 8 (ppm)
Problem 182
UV Spectrum ^max 265 nm (log^c 3.0) ^max 272 nm (log^s 3.1) X-max 294 nm (log10e 1.9) solvent: ethanol
40
80
120
160
200
240
280
m/ e
1H NMR Spectrum (400 MHz, CDCI3 solution)
expansion
expansion
expansion
IT 3.2
T---------- 1— 7.3 7.2
3.0
ppm
,
1
1--
2.6
2.5
ppm
TMS ppm
_JL J
10
1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1--------- 1_____ 1_____ 1_____ I_____ I_____ I_____ I_____ I_____ I_____ I_____ I_____ L.
8
1
0 5 (ppm)
293
Problem 183
1200
1600 2000 V (cm-1)
3000
4000
100
800
Mass Spectrum
80 -■ 03 -. 0C 60 -■ C0/> - 0 . £ 40 7 0 - 0 ^^
^m ax
M+ '= 180
l
UV Spectrum
1£32
^max 330 nm (log^s 4.2) solvent: ethanol
20 »
265 nm (log^s 4.5 )
*
»
*—
1
— *— »—
•— •— •— •— .—
40
80
1
1
ini
— •— •— •— •—
120
•—
160
I 1
— •—
1 —
•—
200
1 —
*—
1 —
240
c 13h 8 o •—
1 —
•—
1 —
•—
1
— •—
280
m/e
i— ■ — i— •— i— 1— i— 1— r 13C NMR Spectrum
expansion
(100.0 MHz, CDCI3 solution)
135
DEPT
125
ppm
CH2^ CH3f C H f -------------expansion
solvent
JL
proton decoupled
i
135
—1------ r-
125
I 1__
ppm
_J__________________ L 2 0 0
160
120
80
40
0
5 (ppm)
8 (ppm)
294
Problem 184
100
Mass Spectrum
45 43
80
29
No significant UV
60
absorption above 220 nm
40
M+’ = 132 (< 1%) 89
20
87 |
117 131
c 6 h 12o 3 ,1
40
80
120
160
200
i
I
I
240
I
I
L_
280
m/e
1— 1— i— 1— r
“i
1
11— r•— r
I
I
I
I
I
I
I
J
I
I
I
[-
I
1
I
13C NMR Spectrum (150.0 MHz, CDCI3 solution)
D EP T
ch 2J c h 3|
CHf
solvent proton decoupled
I
.
I
200
.
I
.
I
160
I
I
I
I
120
I
I
L_
80
40
I
L
0
5 (ppm)
6 (ppm)
295
Problem 185
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e 1
..I
I
'
1
I
I
'
1
I — 1
1
1
^
1
1
1
1
1
'
I
'
I
'
I
i
I 0
i 5 (ppm)
13C NMR Spectrum CDCIj solution)
(100.0 MHz,
D EPT
CH2| CH3f CHf
solvent proton decoupled
i
I
i
I i 200
| I
i
I i 160
I
i
I i 120
I
i
I
I i 80
I
i
I i 40
1H NMR Spectrum (400 MHz, CDCI3 solution)
■ ----
__ _
__
TMS 1
1
10
1
1
1
1
8
1
i
i
1
*
1
1
1
1
I
.
1
1
1
1
1
I
0
5 (ppm)
296
Problem 186
1200
2000
1600 V (cm'1 )
3000
4000
800
No significant UV absorption above 220 nm
120
80
40
200
160
240
280
m/e I
.
I
.
1
I
I
1
I
1
1
'
1
'
1
'
“ 1-----T“
'[
1
1
1
1
1—
4.0
40
80
120
160
200
240
280
m/e I
1
1
' .....1
■
1
'
1
1
T
1
r
...... r.... '
t .... ' 11' "i.....
1
1 1 1
1
1....
1 0
1 8 (ppm )
13C NMR Spectrum (50.0 MHz, CDCI3 solution) two resolved signals at higher field
D EPT
CH2(f CH3f
ch|
expansion
—
two resolved signals at higher field
i
iL
solvent
iii
1
1
.....
8,0
70 ppm
i
proton decoupled
----------- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- «----- 1----- 1----- 1----- 1----- 1----- 1----- 1---- 1------ 1----- 1
200
10
304
9
120
160
8
7
6
80
5
4
1
40
3
2
1
1
0 8 (ppm)
Problem 194
1200
2000
1600 V (cm'1 )
3000
4000
800
100
UV Spectrum
80
lo g 10e > 4.0
60 40
20 120
80
40
160
200
240
280
m/e I
■
I
1
I
■
I
1
I
■»
i
1
• ..i
1
i
i
1
1
'1
'
1
1
'
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EP T
ch 2J
CH3f
ch|
r solvent
proton decoupled ..... l
_____ ______i —... L ._ ....... i i i i i < i i
.
200
10
9
160
8
7
i i
i
i
.
i
120
6
5
.
i
i
i .....i..... j ___ .___ i— .------------
80
4
40
3
2
0
1
5(ppm)
0 6 (ppm)
305
Problem 195
1600 V (cm'1 )
100 : 80 60 40
1200
2000
3000
4000
Mass Spectrum
121/123
•
03
7
Cl
800
41
No significant UV
” 4.0)
M+*
60
227/229
40
197/199
20 C 8 H6 N 0 2 Br 40
80
120
160
200
240
280
m/ e i— 1— r 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EP T
CH2j CH3f
ch|
solvent
Resolves into two signals at higher field
proton decoupled
200
160
120
80
40
0
5 (ppm)
6 (ppm)
327
Problem 217
100 :
91
Mass Spectrum
107
80 • 03 -. 0 Q. 60 -• 0 0 -. -Q 03 40 7 O M + '= 166
20 ! * • *
I, CmHmO? I III,I li .1 j! li I 1,I ............ *—* J-- ■ J-- - -J-- •-- 1-- *-- 1-- •-- 1-- ■-- 1-- CD
.
-Q
7
o
absorption above 220 nm
-
20
102
j*
-I
I I
llll IM il il I I
c 5H9N 0 4
M + ' = 1 4 7 « 1 %)
lL_
I I—
_J
I
I
I____I— I___I
120
80
40
I
I
I
I
I
I
I
200
160
I
I
I
■
I
280
240
m/e i
'
i
r
i
T
r
•
i
1
i
•
1
i
i
•
i
..... I” ..... '
1
1
1
1
..1' 1 '■..... r
....
- ....
13C NMR Spectrum (100.0 MHz, D2 O solution)
D EP T
ch 2J[ ch3| cHf
proton decoupled
----------- 1----- 1----- 1----- 1------1----- 1----- 1----- 1------1----- 1------1------1------1----- j
200
10
378
9
160
8
7
120
6
5
80
4
1
1
1
40
3
2
1
1
0
1
1
8 (ppm)
0 8 (ppm)
Problem 268
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/e
i
i
|
i
r
T
T
T
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH2iJ[ CH3f C H f
solvent
proton decoupled
j
1
i 1 200
i
1
i 1 160
i
i
i 120
.
i
J
80
1
L
40
0
8 (ppm)
5 (ppm)
379
Problem 269
100 7 80
28
1200
2000
1600 V (cm'1 )
3000
4000
Mass Spectrum
56
30
800
*
-. CO (D C 60 “- 0 0 l
.
CO _Q
40
7
O
20
7
No significant UV M+'
absorption above 220 nm
57
c 3 h 7n
Li! 40
80
120
160
200
240
280
m/ e I
1
'
1
'
'
1
'
1
1
1
I
1
'
1
1
I
i.....................................I i i i I i i. iI 80 40
i.
iI
I 0
I
1
I
I
1
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
DEPT
CH2| CH3f CHf
solvent
proton decoupled
.............. ...................................................................................I i i i i I i i i I i
200
10
380
9
160
8
7
120
6
5
4
1
.
3
2
1
i______ 8(ppm)
0 5 (ppm)
.......
I-- ----'---1-- 1
I ' I ' I
Problem 270
IR Spectrum (CHCI3 solution) I
I
4000
.
1
.
«
.
I
.
I
3000
.
I----------------- |-------- 1-------- I--------.--------1--------1-------- 1--------1-------- 1-------- 1-------- 1-------
1200
2000 1600 V (crn1 )
L
800
100 80 No significant UV
60
absorption above 220 nm
40
20 40
80
120
160
200
240
280
m/e
I
!
I
i
I
.
I
I
I
I
I
I
I
I
I
I
| I
I
I
I
I
I
1----1----r
I
I
I
I
13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
CH3f
c h
ch|
solvent proton decoupled
i
»
I
«
I
200
»
I
I
I
160
I
I
I
I
I
120
I
I
80
40
I
0
L
5 (ppm)
1H NMR Spectrum (400 MHz, CDCI 3 solution)
—
J1
A
i— --------r
1.9
1.7
ppm TMS
1
I. ---- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1------- 1 _
10
9
8
7
6
5
1
!
4
1
•
3
1
>
2
1
•
1
1
>
1
0 6 (ppm)
381
Problem 271
4000
3000
1200
2000 1600 V (crrf1 )
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e -i— i— 1— i— 1— r 13C NMR Spectrum (100.0 MHz, CDCI3 solution)
D EPT
ch 2|
CH3f CHf
solvent
proton decoupled
I
10
382
■
I
■
l
8
■
I
I
I
I
I
I
I
I
.
I
I
I
I
I
I
I
I
I
I
1
L_
0 5 (ppm)
4000
3000
2000 1600 V (cm'1 )
1200
800
UV Spectrum
X max
281 nm (log10s 3.5)
X max 230 nm (log10s 3.6) solvent: methanol
40
80
120
160
200
240
280
m/ e 13C NMR Spectrum (50.0 MHz, CDCI3 solution)
D EPT
ch 2|
CH3f CHf
solvent
proton decoupled I
I
I
«
1
160
200
10
1
i _______ I_______ _______ I_______ I_______ I_______ I_______ I_______ I_______ I_______ L_
8
120
80
J
40
i
I
I
I
0
1
L_
5 (ppm)
0 8 (ppm)
383
Problem 273
100
Mass Spectrum
:
43
80 60
" .
CD q .
■ -
(D cf> CD
.
40
ro
-
No significant UV absorption above 220 nm
-Q
7 o - O" '"8
M+
20 :
’=
1 7 5 (< 1 % )
c 6h 9 n o 5
I i
*-----•----- *-----1----- •----- 1----- >-----1----- •-----1----- >----- •----- ■----- 1----- •----- 1----- •----- «----- •----- 1----- •----- 1----- ■----- 1----- •----- «----- •-----1----- 1___________ L
Chapter 10.3 Problems in 2D NMR
m/ e 1
1
1
' .-"T
1
'
1
'
A,(nm) 1
'
1
1
T"” 1
1
'
1
1 ”r
~r.....r .. •
1
1
13C NMR Spectrum (100 MHz, CDCI3 solution)
DEPT
ch 2{
CH3f CHf
solvent proton decoupled
1 — 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1___ 1____
200
160
120
80
40
0 5 (ppm)
6 (ppm)
454
Chapter 10.3 Problems in 2D NMR
1
JL
i
JL
--------- !----------1-------- 1-------- r*
JL
Ju
ppm
JUL
ppm
n----------- 1----------- 1----------- r
C-H Correlation Spectrum (1H 300 MHz; 13C 75 MHz; CDCI3 solution)
$
m
"i 4.0
i
i 3.0
r
-
20
-
40
-
60
n ------------------ 1-------------------1------------------ r
2.0
1.0
ppm
455
Chapter 10.3 Problems in 2D NMR
Problem 319 Use the basic spectral data plus the COSY and C-H correlation spectra on the facing page to deduce the structure of this compound.
4000
2000
3000
1200
1600
800
V (cm-1)
No significant UV absorption above 220 nm
40
120
80
160
200
240
280
m /e
200
5
160
4
120
3
80
40
2
0 8 (ppm)
1
0
8 (PPm)
456
Chapter 10.3 Problems In 2D NMR
I u k I ' 1 1 ' I 1 1 1' I 1 ' 11 I ' ' ' ' I 1 1 1 1 I
ppm
1H - 1H C O S Y Spectrum (500 MHz, CDCI3 solution)
1.0
2.0
4-
*
3.0
4.0
— I 1 1 1 1 I 1 1 1 1 I 1 11 1 I 1 11 1 I 1 1 11 I 1 1 1 1 I 4.0
3.0
2.0
J
1.0
u
J
ppm
l
I 1 " 1 I 1 11 1I
ppm
C-H Correlation Spectrum (1H 500 MHz; 13C 125 MHz; CDCI3 solution)
10
20
30
40
50
60
70
4.0
3.0
2.0
1.0
PPm
457
Chapter 10.3 Problems in 2D NMR
Problem 320 Use the basic spectral data and the NOESY spectrum on the facing page to deduce the structure of this compound.
X (nm)
200
160
80
120 solvent residual
1H NMR Spectrum
40
0 8 (ppm)
exchanges with D 20
*
(4 0 0 M H z, C 6 D 6 solution)
! »
expansion
7 .3
solvent residual
7.1
6 .9 ppm
6 .4
6 .2 ppm
exchanges with D 20
_J____ i
10
458
I___ i
9
L
8
J
L.
1
0 8 (ppm)
Chapter 10.3 Problems in 2D NMR
1H -1H NOESY Spectrum (300 MHz, Benzene-Dg solution) Diagonal peaks plotted with reduced intensity
ppm 3
3 ppm
459
Chapter 10.3 Problems in 2D NMR
Problem 321 Use the basic spectral data plus the COSY and NOESY spectra on the facing page to deduce the structure, including stereochemistry of this compound. 4000
3000
2000 1600 V (cm'1)
1200
800
No significant UV absorption above 220 nm
40
80
120
160
200
240
280
m/ e I
1
I
1
I
1
I
1
I
1
I
1
1
1
'
1
'
1
'
1
'
1
1
1
'
1
I
1
I
1
I
1
1
13C NMR Spectrum (125 MHz, C D C I3 solution)
DEPT
CH 2)|( C H 3f C H f
solvent -ft proton decoupled
. 1 ______, i
1
I
200
1
I
1
I
160
i
I
1
1
120
1
I
1
I
80
1
I
40
0 §(ppm)
6 (ppm)
460
Chapter 10.3 Problems in 2D NMR
A_
-J
A_
Jj
lJ
461
Chapter 10.3 Problems in 2D NMR
Problem 322 Use the basic spectral data plus the COSY and NOESY spectra on the facing page to deduce the structure, including stereochemistry of this compound.
100:
Mass Spectrum 69
80
41
No significant UV
60
absorption above 220 nm
40 M+‘
20
136
jiuL
154
Ci 0 Hi8O
I. _j
40
80
120
160
i
i
i
200
i
■
i
.
240
i
i
i_
280
m/ e
13C NMR Spectrum (125 MHz, CDCI3 solution)
DEPT
CH3t
c h 2|
c h
f
solvent proton decoupled
J
.
I
200
L
_ l
.
I
160
L_
J
120
1
I
I
I
80
I
I
I
I
40
1
I
L-
0 8 (ppm)
5 (ppm)
462
Chapter 10.3 Problems in 2D NMR
A
A
___
A
A
___
ppm
2.0
3.0
4.0
5.0
6.0
463
Chapter 10.3 Problems in 2D NMR
Problem 323 U se the basic spectral data and the C-H Correlation and HMBC spectra on the the following pages to determine the structure of the compound, including its stereochemistry. 4000
100 80 60
■ • . “■ -
3000
2000 1600 V (cm'1)
800
Mass Spectrum
203/205
UV Spectrum log10s >4.5
CD - solvent
r~i- solvent ......J..... proton decoupled
130 _______ __ ___________ I...........
160
i
120
1H NMR Spectrum
-
ppm
I I
1
— i----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1----- 1— i----- 1
200
1 2 0
J.
l
i
i
i
80
i
i
i
i
i
40
i
i
0
§(ppm)
expansion
(400 MHz, Acetone-d6 solution)
J U l
7.4
6.2 ppm
6.6
7.0 expansion scale 0
40 80 120 Hz
solvent residual TMS J _____
10
464
9
8
l
L_
7
6
5
4
3
2
. 1
I
i 0 5 (ppm)
Chapter 10.3 Problems in 2D NMR
C-H Correlation Spectrum solvent residual
( 1H 400 MHz; 13C 100 MHz; Acetone-d6 solution)
JL
u___________________
u
ppm
solvent
40
60
- 80
100
120 -*• -l-i-
- 140
II
ppm
II — rn-------- 1--------- 1--------- 1-------- 1— C-H Correlation Spectrum (expansion) ( 1H 400 MHz; 13C 100 MHz; Acetone-d6 solution)
ppm
00 120
130
00
140
- 150
I 7.4
7.0
6.6
ppm
465
Chapter 10.3 Problems in 2D NMR
C-H HMBC Spectrum ( 1H 400 MHz; 13C 100 MHz; Acetone-d6 solution)
Jl
solvent
466
II
II____________________________________
solvent residual
10.4 NMR SPECTRAL ANALYSIS
46”
Chapter 10.4 NMR Spectral Analysis
Problem 324 Give the number of different chemical environments for the magnetic nuclei 'H and l3C in the following compounds. Assume that any conformational processes are fast on the NMR timescale unless otherwise indicated. Structure CH3 -C O -C H 2CH2CH3 CH 3CH 2 - CO - CH 2CH 3 c h 2= c h c h 2c h 3
cis- CH3CH=CHCH3 frans-CH3CH=CHCH3
0
0
O O ci a: Br—
^ — Br
Br —
Cl
C l^ y O C H s
Assuming slow chairchair interconversion Assuming fast chairchair interconversion H
Assuming the molecule to be
conformationally rigid
468
Number of 1H environments
Number of 13C environments
Chapter 10.4 NMR Spectral Analysis
Problem 325 Draw a schematic (line) representation of the pure first-order spectrum (AMX) corresponding to the following parameters: Frequencies (Hz from TMS):
vA= 300; vM= 240; vx =120.
Coupling constants (Hz):
JAM = 10; JAX = 2; JMX = 8 .
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 6 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
_l
350
300
250
I I I \
-----,-----1-----(-----1-
200
150
100
(Hz from TMS)
469
Chapter 10.4 NMR Spectral Analysis
Problem 326 Draw a schematic (line) representation of the pure first-order spectrum (AMX) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 180; vM= 220; vx = 300.
Coupling constants (Hz):
Jam = 10; JAX = 12 ; J MX = 5.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 8 scale corresponding to the above spectrum obtained with an instrument operating at 200 MHz for protons.
-- 1--- j--- ,--- 1--- 1 ----1--- 1 --- 1----1--- 1--- 1--- 1--- 1----1 --- 1--- 1--- ,--- 1----1--- 1 --- 1--- 1--- 1----1--- 1--- 1 --- 1--- 1----1—
350
470
300
250
200
150
100
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 327 Draw a schematic (line) representation of the pure first-order spectrum (AX2) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 150; vx = 300.
Coupling constants (Hz):
*^AX 2 0 .
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 8 scale corresponding to the above spectrum obtained with an instrument operating at 400 MHz for protons.
-I 1 r 350
-1--- 1--- r 300
250
200
\ 1 1 r 150
_|
100
,
,
f_
(Hz from TMS)
471
Chapter 10.4 NMR Spectral Analysis
Problem 328 A 60 MHz 'H NMR spectrum of diethyl ether is given below. Note that the spectrum is calibrated only in parts per million (ppm) from tetramethylsilane (TMS), i.e. in 5 units. (a)
Assign the signals due to the -C H 2- and -C H 3 groups respectively using three independent criteria (the relative areas of the signals, the multiplicity of each signal and the chemical shift of each signal).
(b)
Obtain the chemical shift of each group in ppm, then convert to Hz at 60 MHz from TMS (see Section 5.5).
(c)
Obtain the value of the first-order coupling constants 3J H.H (in Hz).
(d)
Demonstrate that first-order analysis was justified (see Section 5.9).
TMS
Jm
J
t 3.0
472
2.0
1.0
0.0 8 ppm from TMS
Chapter 10.4 NMR Spectral Analysis
Problem 329 A 100 MHz 'H NMR spectrum of a 3-proton system is given below. (a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (5 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
1H 500
1H 450
1H 400 (Hz from TMS)
473
Chapter 10.4 NMR Spectral Analysis
Problem 330 Portion of the 60 MHz NMR spectrum 2-furoic acid in CDCI3 is shown below. Only the resonances due to the three aromatic protons (Ha, Hm and Hx) are shown.
/ Ha "
\ O
2-Furoic Acid COOH
(a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (8 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
Note: This is a 60 MHz spectrum.
H„
474
H„
Chapter 10.4 NMR Spectral Analysis
Problem 331 COOH
A portion of the 100 MHz 'H NMR spectrum of 2-amino-5-chlorobenzoic acid in CD3OD is given below. Only the resonances due to the three aromatic protons are shown.
H4
(a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (5 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
(c)
Assign the three multiplets to H3, H4 and H 6 given: •
the characteristic ranges for coupling constants between aromatic protons (see Section 5.9);
•
the fact that H3 will give rise to the resonance at the highest field due to the strong influence of the amino group (see Table 5.6).
475
Chapter 10.4 NMR Spectral Analysis
Problem 332a Portion of 100 MHz 'H NMR spectrum of methyl acrylate (5% in CeDg) is given below. Only the part of the spectrum containing the resonances of the olefinic protons Ha, H b and Hc is shown.
/ , . / C = C V. Hb He methyl acrylate
476
(a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (5 in ppm) by direct measurement.
(c)
Justify the statement that "this spectrum is really a borderline second-order (strongly coupled) case". Point out the most conspicuous deviation from first-order character in this spectrum (see Section 5.9).
(d)
Assign the three multiplets to HA, HB and Hc on the basis of coupling constants only (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 332b This is the computer-simulated spectrum corresponding to the complete analysis of the spectrum shown in Problem 318a, i.e. an exact analysis in which first-order assumptions were not made. The simulated spectrum fits the experimental spectrum, verifying that the analysis was correct. Compare your (first-order) results from Problem 3 18a with the actual solution given here. Number of SPINS F(l) F(2) F(3) J(l,2) J(l,3) J(2,3)
3 = + 528.500 Hz = + 594.531 Hz = + 626.093 Hz + 10.539 Hz + 1.589 Hz + 17.278 Hz
START of simulation = + 750.000 Hz FINISH of simulation = + 500.000 Hz LINE WIDTH = + 0.427 Hz
Chapter 10.4 NMR Spectral Analysis
Problem 333 Draw a schematic (line) representation of the pure first-order spectrum (AX3) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 160; vx = 280.
Coupling constants (Hz):
JAX =15.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 5 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
I
350
478
I
I
I
I
I
300
I
I
I
I
I
250
I
I
I
I
I
200
I
I
I
I
I
150
I
I
I
I
I
100
I
I
I
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 334 A 100 MHz *H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
U l 3H 200
1H 150
100
(Hz from TMS)
479
Chapter 10.4 NMR Spectral Analysis
Problem 335 Draw a schematic (line) representation of the pure first-order spectrum (AMX2) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 340; vM= 240; vx = 100.
Coupling constants (Hz):
J AM = 10; JAX = 2; J MX = 6 .
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 5 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
i
350
480
1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-
300
250
200
150
100
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 336 Draw a schematic (line) representation of the pure first-order spectrum (AM2X) corresponding to the following parameters: Frequencies (Hz from TMS):
vA = 110; vM= 200; vx = 290.
Coupling constants (Hz):
JAM = 10; J AX = 12; J MX = 3.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 8 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
I
350
I
I
I
“I
I
300
I
I
I
I
I
250
I
I
I
1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1-------1—
200
150
100
(Hz from TMS)
481
Chapter 10.4 NMR Spectral Analysis
Problem 337 A 100 MHz 'H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (5 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
I 2H
1H 600
482
500
1H 400
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 338 A 100 MHz 'H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 339 A 100 MHz 'H NMR spectrum of a 4-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 340 A 200 MHz 'H NMR spectrum of a 5-proton system is given below. (a)
Draw a splitting diagram.
(b)
Analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(c)
Justify the use of first-order analysis (see Section 5.9).
2H
2H 1010
1H
— r~
—i—
990
Hz from TMS
510
490 Hz from TMS
90
110 Hz from TMS
485
Chapter 10.4 NMR Spectral Analysis
Problem 341 COOH
Portion of 100 MHz NMR spectrum of crotonic acid in CDCI3 is given below. The upfield part of the spectrum, which is due to the methyl group, is less amplified to fit the page.
Ch/
c=c \ Hm crotonic acid
(a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (Jin Hz) and chemical shifts (5 in ppm) by direct measurement. Justify the use of first-order analysis.
(b)
There are certain conventions used for naming spin-systems {e.g. A M X , AMX2, AM2X3). Note that this is a 5-spin system and name the spin system responsible for this spectrum (see Section 5.9).
730
720
710
700
690
600
590
580
(Hz from TMS)
486
570
200
190
180
Chapter 10.4 NMR Spectral Analysis
Problem 342 The 100 MHz 'H NMR spectrum (5% in CDC13 ) of an a,(3-unsaturated aldehyde C4H60 is given below. (a)
Draw a splitting diagram and analyse this spectrum by first-order methods, i.e. extract all relevant coupling constants (J in Hz) and chemical shifts (8 in ppm) by direct measurement.
(b)
Justify the use of a first-order analysis (see Section 5.9).
(c)
Use the coupling constants to obtain the structure of the compound, including the stereochemistry about the double bond (see Section 5.9).
Chapter 10.4 NMR Spectral Analysis
Problem 343 Draw a schematic (line) representation of the pure first-order spectrum (AMX3) corresponding to the following parameters: Frequencies (Hz from TMS):
vA= 80; vM= 220; vx = 320.
Coupling constants (Hz):
JAM = 10; J AX = 12; J MX = 0.
Assume that the spectrum is a pure first-order spectrum and ignore small distortions in relative intensities of lines that would be apparent in a "real" spectrum. (a)
Sketch in "splitting diagrams" above the schematic spectrum to indicate which splittings correspond to which coupling constants.
(b)
Give the chemical shifts on the 5 scale corresponding to the above spectrum obtained with an instrument operating at 60 MHz for protons.
1
350
488
1
1
1
1
1
300
1
1
1
1
1
250
1
1
1
1
1
200
1
1
1
1
1
150
1
1
1
1
1
100
1
1
1-
(Hz from TMS)
Chapter 10.4 NMR Spectral Analysis
Problem 344 A portion of the 90 MHz *H NMR spectrum (5% in CDCI3) of one of the six possible isomeric dibromoanilines is given below. Only the resonances of the aromatic protons are shown. Br
Br Br
Br
NH2
NHo
nh2
NH2
Br
Br Br
Br
Determine which is the correct structure for this compound using arguments based on symmetry and the magnitudes of spin-spin coupling constants (see Section 5.9).
4^9
Chapter 10.4 NMR Spectral Analysis
Problem 345 The 400 MHz 'H NMR spectrum (5% in CDCI3 after D2O exchange) of one of the six possible isomeric hydroxycinnamic acids is given below. OH
OH
OH
Determine which is the correct structure for this compound using arguments based on symmetry and the magnitudes of spin-spin coupling constants (see Section 5.9).
1H
490
Chapter 10.4 NMR Spectral Analysis
Problem 346 In a published paper, the 90 MHz 'H NMR spectrum given below was assigned to 1,5-dichloronaphthalene, C,„H6C12. 1,5-dichloronaphthalene
(a)
Why can't this spectrum belong to 1,5-dichloronaphthalene?
(b)
Suggest two alternative dichloronaphthalenes that would have structures consistent with the spectrum given.
491
Chapter 10.4 NMR Spectral Analysis
Subject Index
Subject Index Key:
13C NM R = Carbon 13 nuclear magnetic resonance spectroscopy NM R = Proton nuclear magnetic resonance spectroscopy 2D NM R = 2-dimensional NMR IR = Infrared spectroscopy MS = Mass spectrometry UV = Ultraviolet spectroscopy
Absorbance, molar
8, 9
Aldehydes
Anisotropy, magnetic, NM R
50,51
Appearance potential, MS
21
13C NMR
76
'H NM R
45, 46
,3C NM R
76, 80
IR
17,18
coupling constants, NMR
68
'H N M R
4 5 ,4 9 ,5 0
Aromatic compounds
Alkanes l3C NM R
76, 78
polynuclear
4 9 ,80
'H NMR
45, 46, 66
UV
13
Alkenes
Aromatic Solvent Induced Shift
94
l3C N M R
76,78
(ASIS)
'H N M R
4 5 ,4 8 ,6 7
Auxochrome,
10, 13
IR
19,20
Base peak, MS
24
UV
11
Bathochromic shift, UV
10
Beer-Lambert Law
2, 8
Alkynes l3C NMR
76, 79
Boltzmann excess, NM R
36
'H N M R
45 ,4 8
Cation radical, MS
21,22
IR
19,20
Carbonyl compounds
Allenes
i3C N M R
76-80
l3C N M R
75
IR
17,18
IR
20
MS
32,33
UV
12
Amides l3C NMR
76
'H N M R
4 6 ,5 1 ,9 8
l3C NM R
76, 77
IR
18
'H N M R
45,51
IR
18
MS
33
Amines 'H N M R
4 6 ,5 1 ,9 7
IR
19
Analysis o f 'H NMR Spectra Anion Radical, MS
Carboxylic acids
Chemical Ionisation, MS
22
56-65 21
Organic Structures fro m Spectra, Fifth Edition. L. D. Field, S. Stemhell and J. R. Kalman. © 2013 John Wiley & Sons, Ltd. Published 2013 by John Wiley & Sons, Ltd.
493
Subject Index
Chemical shift aromatic solvent induced
42 94
(ASIS)
i3C N M R
76
IR
18
MS
32,33
l3C, tables
75-80
factors influencing
42-44,
Exchange broadening, NM R
96
50, 51
Exchangeable protons, NM R
51, 52, 97
]H, tables
45-49
l9F N M R
35,99
scale
43 ,44
First-order spectra, NMR
57
standard
42,43
rules for analysis
Chirality, effect on NM R
98,99
Fourier transformation, NM R
40
Chromophore
3
Fourier Transform Infrared, FTIR
16
Fragmentation, MS
21,26-33
Cleavage, MS
56, 59-60
a-
32
common fragments
27
p-
31
Free induction decay (FID), NMR
40
Conformational exchange
97,98
Halogen derivatives, IR
19
Connectivity
4
Halogen derivatives, isotopes, MS
27-29
Contour plot, 2D NMR
82
Heteroaromatic compounds
Correlation Spectroscopy (COSY)
83,84
2D NM R
i3C N M R
80
'HNMR
49
Heteronuclear Single Quantum
Coupling constant
494
Esters,
82, 84-85
NMR
53
Correlation (HSQC), 2D NMR
allylic
67
Heteronuclear Shift Correlation
aromatic systems
68-70
(HSC), 2D NMR
geminal
66
High-resolution mass spectroscopy
24-25
heteroaromatic systems
68
Hydrogen bonding, IR
17
long range
67,88
Hydroxyl groups
olefmic
67
IR
17
vicinal
66
'H N M R
4 5 ,5 1 ,9 7
Cyanates, IR
20
Hypsochromic shift, UV
10
Degree o f Unsaturation
3 ,4
Imines, IR
19
Deshielding, NMR
4 4 ,5 0 ,5 1
Intermolecular exchange
97
Dienes, UV
11
Ionisation, MS
D20 exchange
51,52
chemical ionisation (Cl)
22
DEPT, l3C NMR
72,73
electron impact, (El)
21-23
Electrospray Ionisation (ESI), MS
22
electrospray (ESI)
22
Enol ethers, IR
19
matrix assisted (MALDI)
22
Equivalence, NMR
4 3 ,57
Isotope ratio, MS
27-29
accidental
43
Isocyanates, IR
20
chemical
4 3 ,57
Karplus relationship, NM R
66
magnetic
57
82, 84-85
Subject Index
Ketones
Saturation, NM R
37
13C NMR
76
Sensitivity
5
IR
18
Shielding, NM R
4 3 ,5 0 ,5 1
MS
32,33
Solvents for NM R
95
4 5 ,5 1 ,9 7
Spectrometry, Mass
21
Labile protons
Spectroscopy, definition of
Lactones
1
IR
18
IR
15
Larmor equation
35
NM R
34
M + 1, M+2 peaks, MS
27-28
,3C N M R
71
Magnetic anisotropy
50-51
continuous wave (CW)
39
M cLafferty rearrangement
33
Fourier transform (FT)
40
MALDI, MS
22
UV
7
Mass number, MS
24
pH dependence
13-14
Mass spectrometry
21
solvent dependence
14
Matrix Assisted Laser Desorption
22
Ionisation, (MALDI) MS
Spin, nuclear, NM R
34
Spin decoupling, NMR
65
Metastable peaks, MS
29
broadband
71
Molecular ion, MS
21
noise
71
Nitrogen Rule, MS
24
selective
65
Nitriles
Spin quantum number, NMR
34, 35
13C NMR
76-78,80
'H N M R
46-49
strongly coupled systems
56,57
IR
20
weakly coupled systems
56,57
Spin-spin coupling
53
Nitro compounds, IR
19
NM R spectroscopy
34,71
NM R time-scale
96-98
Splitting diagram, NM R
61
Nuclear Overhauser effect (NOE)
41
Structural element
3
Sulfonamides, IR
19
NM R
Spin system, NM R naming conventions
56 57
NOESY, 2D NMR
82,91-92
Sulfonate esters, IR
19
31p n m r
35,99
Sulfones, IR
19
Partial double bonds
98
Sulfoxides, IR
19
Time o f Flight (TOF), MS
24
Polynuclear aromatic compounds 13c n m r
80
Two-dimensional NMR
81
'h n m r
4 6 ,4 9 ,6 8
T 1? NMR
37
Prochiral centre
98-99
Thiocyanates, IR
20
Relaxation, NM R
37
Thiols, 'H N M R
4 5 ,5 1 ,9 7
spin lattice
37
TOCSY, 2D NM R
82,92-93
Residual solvent peaks
94-95
Wavenumber, IR
15
Resonance, NM R
35
Ring current effect, NM R
50
495
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